50 amc lectures chapter 33 logarithms basic … amc lectures chapter 33 logarithms basic knowledge....

50 AMC Lectures Chapter 33 Logarithms

BASIC KNOWLEDGE 1. Definition If a and x are positive real numbers and a ≠ 1, then y is the logarithm to the base a of x. This can be written as

y = loga x (1.1)

if and only if x = a y (1.2) A quick way to convert the logarithm to the exponent:

Example 1: Rewrite the logarithmic forms as the exponential form: (1): log5 x = y ⇒ 5y = x (2): log53 = y ⇒ 5y = 3

(3): loga1024 = 10 ⇒ a10 = 1024. (4): log10 1000 = 3 ⇒ 103 = 1000. Notes: (1) The logarithm base 10 is called the common logarithm. log x always refers to log base 10, i.e., log x = log10 x. (2) The logarithm base e is called the natural logarithm. ln x = loge x. (3) log10 x is always written as log x. (4) Zero and negative numbers have no logarithm expressions. You will not be able to see any expressions like log a 0 , loga (− 3), or . x10log−

The graphs of logarithmic functions f (x) = log a x all have x-intercept 1, and are increasing when a > 1 and decreasing when a < 1.

228


y = loga x (a > 1) y = loga x (a < 1)

Example 2: (1970 AMC) If and 225log8=a ,15log2=b then

(A) 2ba = (B)

32ba = (C) a = b (D)

2ab = (D)

23ba =

Solution: (B). The exponential form of the given equations is 8a = 225 and 2b = 15.

Since 8 = 23 and 152 = 225, we have (23)a = 225 = 22b, so 3a = 2b and 32ba = .

Example 3: (1974 AMC) If p=3log8 and ,5log3 q= then, in terms of p and q, equals

5log10

(A) pq (B) 5

3 qp + (C) qppq

++ 31 (D)

pqpq

+13

Solution: (D). The exponential form of the given equations is 3 = 8p = 23p and 5 = 3p. 5 = (23p)q = 23pq. We are asked to find i.e. x such that 10x = 5. Since 5 = 23pq, we have 10x = 23pq.

,5log10 x=

10x = 2x · 5x = 2x · 23pqx = 2x(1+3pq). It follows that

x(1 + 3pq) = 3pq and x = pq

pq+13 .

2. Properties of Logarithms: If a, b, x and y are positive real numbers, a ≠ 1, b ≠ 1, and r is any real number, we have

229


230

0

1

xr

1log =a (2.1)

log =aa (2.2)

a xa =log (2.3) ar

a =log (2.4)

loga b log b c = log a c (2.5) 3. The Laws of Logarithms: Law 1: the Product Identity

loga xy = log a x + log a y (3.1) Proof : Let m = loga x and n = loga y. The exponential form of the equations is am = x and an = y. Multiplying these two equations, we get aman = xy ⇒ am +n = xy Transforming the exponential form to the logarithmic form, we get loga xy = m + n.

Since m = loga x and n = loga y, loga xy = log a x + log a y. Example 4: Calculate:

2log2log3log)3(log 6662

6 +⋅+

Solution: 1. 2log2log3log)3(log 666

26 +⋅+ 2log)2log3(log3log 6666 ++=

2log3log2log)23(log3log 66666 +=+×= .1)23(log6 =×= Law 2: the Quotient Identity

x = loga x − loga y (3.2) loga (

y)

Proof: Let m = loga x and n = loga y. The exponential form of the equations is am = x and an = y Dividing these two equations, we get am / an = x/y ⇒ am − n = x/y


Use the definition of the logarithm, we get loga (xy

) = m − n.

Since m = loga x and n = loga y, loga (xy

) = loga x − loga y.

Example 5: (1967 AMC) If x is real and positive and grows beyond all bounds, then

approaches: )12(log)56(log 33 +−− xx(A) 0 (B) 1 (C) 3 (D) 4 (E) no finete nimber Solution: (B). For real x > 1, 6x – 5 and 2x + 1 are positive, so both logarithms are defined. Now

)12(log)56(log 33 +−− xx )12

83(log12

836log1256log 333 +

−=+

−+=

+−

=xx

xxx approaches

as x increases beyond all bounds, because 1log3 =12

8+x

then approaches 0.

Law 3: the Power Identity Formula 1: loga xr

= r loga x (3.3) Proof: Since , x rx

ar

aaax )(loglog log=aax log= →loga x r = (r loga x)loga a.

Recall that . Substituting this value into the equation, we have

.

1

2

log =aa

xr alog=xralog

Example 6: Which pairs are the same?

(A) , (B) log xy = .log2 xy = ,log xy = .log2

xxy =

(C) y = x. (D) , ,10log xy = )1log( 2 −= xy )1log()1log( −++= xxy . Solution: (B).

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The domain is the same (0, +∞) for ,log xy = or xxy

2

log= .

For x ∈ (0, +∞), xxx

=2

. Therefore, xy log= and xxy

2

log= are the same.

232

Formula 2: bmnb a

nam loglog =

c

(a, b > 0, a ≠ 1) (3.4)

Proof: Let . bn

am =log

The exponential form of the equations is ncm ba =)( .

Taking the logarithm of both sides yields or bnac am

a loglog = bncm alog= .

bmnc alog= ⇒ b

mnb a

nam loglog = .

Example 7: Calculate )2log4log8)(log5log25log125(log 525125842 ++++ . Solution: 13.

We use the formula bmnb a

nam loglog to simplify the given equation: =

)2log2log2)(log5log315log5log3( 555222 ++++

.135log5log132log35log

313

2

252 ==⋅=

4. Change-of-Base Theorem:

Formula 1: loga x =logb xlogb a

(4.1)

Proof: Let y = loga x. The exponential form of this equation is ay = x. Taking the logarithm on both sides, we get logb ay = logb x ⇒ ylogb a = logb x

Dividing both sides by logb b, we obtain y =logb xlogb a

⇒ loga x =logb xlogb a

.


Example 8: (1974 AMC) If p=3log8 and ,5log3 q= then, in terms of p and q, equals

5log10

(A) pq (B) 5

3 qp + (C) qppq

++ 31 (D)

pqpq

+13 (E) p2 + q2

Solution: (D). Since we need to find a logarithm in base 10, we must first convert all the given information into that base. This can be done by using (4.1).

From 8log3log3log

10

108 ==p , and

315log5log

10

103 og

q == , we obtain ,8log3log 1010 p=

thus ,3log5log 1010 q= 3

101010 )5

10(log8log5log pqpq == ).5log1( 103 −= pq

Soving this for gives choice (D) as the answer. 5log10

Formula 2: m ≠ 0 (4.2) ,loglog m

aa NN m= Proof:

.logloglog

loglog

logloglog m

amb

mb

b

b

b

ba N

aN

amNm

aNN m====

Formula 3: ,log

1loga

NN

a = N ≠ 1 (4.3)

Proof:

aaNN

NN

Na log

1logloglog == .

Example 9: (1974 AMC). If p=3log8 and ,5log3 q= then, in terms of p and q, equals

5log10

(A) pq (B) 5

3 qp + (C) qppq

++ 31 (D)

pqpq

+13 (E) p2 + q2

Solution: (D).

233


We know that a

bb

a log1log = and can write ,

2log31

2log1

8log13log

33

338 ====p

so ,312log3 p

= and .31

3

315log2log10log

5log5log333

310 pq

pq

qp

qq+

=+

=+

==

234

Formula 4: N

N aa1loglog −= (4.4)

Proof:

NNN aaa

1log)1(loglog 1 −== −

Formula 5: NN

aa 1loglog −= (4.5)

Proof:

NNNa

aa 11 logloglog 1 −== −

−

Formula 6: m ≠ 0 (4.6) ,loglog NmN maa = Proof:

NmNN mm am

aa logloglog == Example 10: (2009 ARML) Compute all real values of x such that

. )(loglog)(loglog 4422 xx = Solution: By the formula (4.6), we have =)(loglog 22 x 2

4444 )log2(log)log2(log2 xx = .

Therefore = ⇒ ⇒

⇒

244 )log2(log x

0log) 42 =− x

)(loglog 44 x

(log4

xx 42

4 log)log2( =

0)(log4 4 x 1log4)( 4 =−xx

We have ⇒ (extraneous). 0log4 =x 140 ==x

Or ⇒ 01log4 2 =−x41log4 =x ⇒ 24 4/1 ==x .


235

Formula 7: NM

NM

b

b

a

a

loglog

loglog

= (a > 0, a ≠ 1, b > 0, b ≠ 1, M, N > 0, N ≠ 1) (4.7)

Proof :

By the Change-of-Base Theorem, we have MNM

Na

a logloglog

= , and MNM

Nb

b logloglog

= .

Therefore NM

NM

b

b

a

a

loglog

loglog

= .

Nn

N an

a log1log =Formula 8: , N is integer greater than 1. (4.8)

Example 11: Find if 8log 3 if a=3log12 .

Solution: 3(1− a

a)

.

)3log12(log33

12log34log364log8log 333333 −====

= a

aa

)1(3)11(3)13log

1(312

−=−=− .

5. Solving Logarithm Equations If x, y and a ≠ 1 are positive real numbers, x = y if and only if loga x = loga y. Similarly xa = ya if and only if x = y. When we use the laws of logarithms, we need to be careful about the following restrictions.

When we apply Law 1 the Product Identity loga xy = log a x + log a y or Law 2: the

Quotient Identity lx

oga (y

) = loga x − loga y, we need to know that both x and y in the

expressions loga xy and loga (xy

), can be negative. However, in the expressions log a x +

log a y or loga x − loga y, neither of them can be negative.


Example 12: (2001 ARML Tiebreak #3) Given 21

)log(

)log(=

yxxy , increasing y by 50%

decreases x by a factor of k. Computer k. Solution: Method 1 (Official solution):

Step 1:yx

yxxy log)log(

21)log( == (1)

Step 2: yxxy = (2)

Step 3: 23

1

yx = (3)

Step 4: 31y

x = (4)

Increasing by 50% gives xyy 27

81278

)23(

13

3=⎟⎟

⎠

⎞⎜⎜⎝

⎛⋅= . Thus

278

=k .

Note: The given equation is valid for x, y < 0. From step 2 to step 3 in the official

solution, they separated yx to

yx , which is incorrect for the case x, y < 0.

The correct way should be: Squaring both sides of (2): yxxy =2)( ⇒ 3

1y

=x .

Method 2 (Solution given by Intermediate Algebra from Art of Problem Solving): Multiplying both sides by 2 log (x/y), the equation becomes

yxxy log)log(2 = (1)

So (2) yxyx logloglog2log2 −=+Rearranging this gives 0log3log =+ yx (3) Applying logarithm identities to the left sides gives

236


)log(logloglog3log 33 xyyxyx =+=+ , so , which mean . Hence if y increases by 50%, meaning that y is multiplied by 3/2, then x must be multiplied by a factor of (2/3)3 = 8/27.

0)log( 3 =xy 13 =xy

Note: The given equation is valid for x, y < 0. From equation (1) to equation (2), we think that the author made a mistake by applying the product identity. We cannot apply the product identity because (2) is not valid when x, y < 0. When we apply Law 3: the Power Identity loga xr

= r loga x, we need to be sure that in

the expressions loga xr, x can be negative if r is even. However, x must be positive in the

expression r loga x. Example 13: (2001 ARML) Compute the largest real value of b such that the solutions to the following equation are integers: . 4

222

2 1010 log)(log xx b =

Solution: Method 1 (from the book "Intermediate Algebra" by Art of Problem Solving): We remove the exponents from the arguments, x2b and x4:

xxb 1010 22

2 log4)log2( = (1)

Noting that xxx 210/1

22 log101loglog 10 == , our expression becomes:

xxb2

22 log

52)log

5( = (2)

Multiplying both sides of (2) by 25 gives us: xxb 2

22

2 log10)(log = ⇒ . 0

2b

log)10log( 222 =− xxb

We have or . 0log2 =x 010log22 =−xb

The first equation gives us x = 1 no matter what b is. The second equation gives us

. /102x =In order for x to be an integer, the expression 10/b2 must be a positive integer. The largest value is 10=b .

237


Note: We believe that equation (1) is incorrect, since you cannot apply the power identity. The equation is true for x < 0, however (1) will not be

true, so (1) is not equivalent to the given equation.

42

222 1010 log)(log xx b =

Method 2 (Official solution):

xx b1010 2

222

log4)(log = (1)

⇒ (2) 0log4)(log4 1010 22

22 =− xxb

0)1)(log)((log 1010 22

2 =−xbx

238

0

0

If , then x = 1 . log 102=x

If , then 1log 1022 =−xb 22

1log 10 bx = , giving 22

10110 2)2( bbx == .

There are several values that b can hold so that x an integer. If 1±=b , the . If 102=x

310

±=b , the . But the largest value of b such that the second solution an integer

is

92=x

10=b , which gives us x = 2. Note: We believe that both equations (1) and (2) are incorrect. The given equation is valid whether or not x > 0 or x < 0 but (2) and (3) will not be true if x < 0, so (2) and (3) are not equivalent to the given equation. Method 3 (our solution): The correct way should be the following:

42

222 1010 log)(log xx b = ⇒ (We should make sure

that the argument is still positive even when x < 0).

22

222 1010 log2)log( xxb =

Taking b2 out, the equation becomes ⇒ 2

222

22

1010 log2)(log xxb = 0log2)(log 22

222

21010 =− xxb

⇒ . 0log)2log( 22

22

21010 =− xxb

This gives us or . 0log 2210 =x 02log 2

22

10 =−xb


The first equation results in . Solving for x, we get x = 1 or x = − 1. We can see that both values of x are the solutions to the given equation. Under these values of x, b can be any value. So there is no largest value of b possible.

12 =x

Solving the second equation, we get . 2log 22

210 =xb

Since we want the largest value of b and we can assume that b ≠ 0, we have

22

2

2log 10 bx = ⇒ 22

202102 2)2( bbx ==

Therefore 2210

2120

1 2)2( bbx == or 2210

2120

2 2)2( bbx −=−= . In order for x to be an integer, the expression 10/b2 must be a positive integer. Since b is real number, there are many values that b can hold such that x is an integer for example,

, 1±=b 2±=b , 5±=b , n

b 10±= , where n can be any positive integer. The largest

such value is 10=b . Example 14: Solve )12(log2log 2

11 −+= −− xx xxx

Solution: By Law 3: the Power Identity, loga xr

= r loga x, the given equation can be written as )12(log2log 2

11 −+= −− xxx

xx

1

(1)

So )12(2 2 −+= xxx

2 =x ⇒ x = 1 or x = – 1. x = 1 is extraneous. We obtained an extraneous solution because we missed something when using the law 3. Note that law 3 has three restrictions: 1 – x > 0, 1 – x ≠ 1, and or x < 1. With these restrictions, we can rule out the extraneous solution x = 1.

012 2 >−+ xx

Example 15: Find the value of y if

⎪⎩

⎪⎨⎧

=+−+

=−+

(2) .021log)2(log2

(1) ,0)22(log

55

25

yx

xx.

239


Solution: Simplifying (1), we get x2 + 2x – 2 = 1. Solving this quadratic for x, we get x = 1, x = – 3. x = – 3 is extraneous, since it yields a negative value that we must take the logarithm of when we substitute x into equation (2).

Simplifying (2), we get 1)2(5 2

=+

yx .

Substituting x = 1 into the above equation yields 59)21(5 2 =+=y . 6. Logarithm Applications Example 16: Find log616 if log1227 = a.

Solution: 4(3 − a)

3+ a.

Method 1:

log616 = )312(log

312log4

36log21

4log26log

16log

12

12

12

12

12

12

×==

a

a

311

)311(4

27log311

)27log311(4

3log1)3log1(4

12

12

12

12

+

−=

+

−=

+−

= .3

)3(4aa

+−

=

Method 2:

3log14

6log2log42log416log

22

266 +

=== (1)

a=+

=×

=3log2

3log3)32(log

3log27log2

2

2

2

3

212 ⇒

aa

−=

323log2 .

aa

aa +

−=

−+

=3

)3(4

321

416log6 .

Example 17: Find (2x)x if 2444 1 =− −xx . (A) .55 (B) 25. (C) .525 (D) 125.

240


Solution: (C).

2444 1 =− −xx ⇒ 244414 =− xx ⇒ 4x = 32

∴ 22122log42log32log 4

244 +=+=×==x .

2x = 5 ⇒ 5255)2( 212

==+xx .

Example 18: Find the product of α and β if they are two distinct roots of the equation (log 3x)(log 5x) = k.

(A) log3 · log5 – k. (B) log15. (C) .151 (D) 15.

Solution: 151

=αβ .

Since α and β are two distinct roots, we have k=⋅ αα 5log3log (1) k=⋅ ββ 5log3log (2) (1) – (2): ββαα 5log3log5log3log ⋅−⋅ = 0.

.0)log5)(loglog3(log)log5)(loglog3(log =++−++ ββαα

.0)log(log)log(log15log 22 =−+− βαβα 0)]log(log15)[(loglog(log =++− βαβα

Since α ≠ β, and log α – log β ≠ 0, log 15 + (log α + log β) = 0 ⇒ log 15 = – log αβ

Therefore 151

=αβ .

Example 19: Find if log xabcd , log mxa = , log nxb = , log pxc = qxd =log . x ≠ 1.

Solution: mnpq

mnp + mpq + mnq + npq.

dcbaabcdx

xxxxxabcd loglogloglog

1log

1log+++

==

241


xxxx dcba log1

log1

log1

log1

1

+++= .1111

1npqmnqmpqmnp

mnpq

qpnm+++

=+++

=

Example 20: If a, b, c form a geometric sequence and form an arithmetic sequence, find the common difference of the arithmetic sequence.

,log ac ,log cb log ba

Solution: 3/2. Let . dbcda abc −==+ logloglog

dab

bcd

ca

−==+loglog

loglog

loglog

bc

aadb

ccda

loglog

logloglog

logloglog

=−

=+

∴ bc

acadbcda

loglog

loglogloglogloglog

=+

−++ .

(1) loglog

log

loglog

bc

acacdab

=+

Substituting b2 = ac into (1) yields 2loglog2loglog ccacdab ==+ .

Therefore 23

)log(

)log(

log

log

log

log 2322

====

ac

ac

acaca

c

ac

abc

d .

Example 21: Find the maximum value of xx −+ 7loglog if – 1 ≤ x ≤ 2, x ≠ 0.

Solution: 1.

449)

27(log7log7loglog 22 −−=−=−+ xxxxx .

We know that – 1 ≤ x ≤ 2, 23

27

29

−≤−≤− x , 481)

27(

49 2 ≤−≤ x .

8449)

27(10 2 ≤−−≤− x .

242


∴ 10449)

27( 2 ≤−−x .

Equality occurs when x = 2. Therefore the greatest value of xx −+ 7loglog is log 10 = 1.

Example 22: Find a/b if baba loglog)](21log[2 +=− .

Solution: 223 +=ba

For log a, log b, and log[12

(a − b)] to be defined, the following statement must be true: a

> b > 0.

The given equation can be written as )log()](21log[ 2 abba =− .

∴ abba =− 2)](21[ ⇒ a2 – 6ab + b2 = 0 .

Since that b > 0, we can divide both sides of the equation by b2 and obtain:

0162

=+⎟⎠⎞

⎜⎝⎛−⎟

⎠⎞

⎜⎝⎛

ba

ba .

Solving for a/b, we get 223 +=ba or 223 −=

ba . The latter answer is extraneous

since 1223 <− . Example 23: (1983 AIME) Let x, y and z all exceed 1 and let w be a positive number such that and , 40log =wy24log =wx .12log =wxyz Find .log wz

Solution: 60. Method 1: Changing the given logarithms into into exponential forms, we get

, 40 wy =24 wx = , wxyz =12) ( .

51

103

211212

12 www

wyx

wz === .

Therefore and . 60zw = 60log =wz

243


Method 2:

244

tLet . We then have wz =logt

wz loglog = , ,24

loglog wx = and40

loglog wy = .

12log40

log24

loglog

logloglogloglog =

++=

++=

twww

wzyx

wwxyz

Or 1211

401

241

=++t

.

Solving for t, we get t = 60. Thus 60log =wz . Method 3:

.601

401

241

121logloglogloglog =−−=−−== yxxyz

xyzxyz wwwww

∴ . 60log =wz

Method 4:

By (4.3), we have ⇒ 12log =wxyz 121log =xyzw (1)

By (4.1): (1) becomes 121

loglog

loglog

loglog

=++wz

wy

wx ⇒

121

log1

log1

log1

=++www zyx

601

401

241

121

log1

=−−=wz

⇒ 60log =wz .

Example 24: (1984 AIME) Determine the value of ab if and

5loglog 248 =+ ba

.7loglog 248 =+ ab

Solution: 512. Method 1: Adding the two given equations, we get:

12loglog 2248 =+ baab

By the Change-of-Base formula, we can change the two logarithms from base 8 and 4,

respectively into base 2: 12log3

log2

2 =+ abab .


So ,12 9log2 =ablog34

2 =ab and ab = 29 = 512.

Method 2:

22

2488 loglogloglog baba +++ = 12loglog 22

48 =+ baab

22

332

)(log)(log 23 abab += 34

223

2 )(logloglog ababab =+=

Therefore 12)(log 34

2 =ab ⇒ 1234

2)( =ab ⇒ ab = 29 = 512.

245


PROBLEMS

Problem 1: (1967 AMC) Given ,log)(log)(log)(log xr

cq

bp

a=== all logarithms to the

same base and x ≠ 1. If yxacb

=2

, then y is:

(A) rp

q+

2

(B) q

rp2+ (C) 2q – p – r (D) 2q – pr (E) q2 – pr

Problem 2: (1966 AMC) If ,loglog MN NM = M ≠ N, MN > 0, M ≠ 1, N ≠ 1, then MN

equals: (A) 21 (B) 1 (C) 2 (D) 10 (E) a number greater than 2 and less than 10.

Problem 3: Simplify .32log)2log2)(log3log3(log 4

29384 −++ Problem 4: Solve 325loglog2 25 =+ xx . Problem 5: (1971 AMC) If ))(log(loglog))(log(loglog 243432 yx =

,0))(log(loglog 324 == z then the sum x + y + z is equal to (A) 50 (B) 58 (C) 89 (D) 111 (E) 1296 Problem 6: Find m if 16loglog8log4log 4843 =⋅⋅ m .

(A) .29 (B) 9. (C) 18. (D) 27

Problem 7: (1996 ARML) For integers x an y with 1 < x, y ≤ 100, compute the number of ordered pairs (x, y) such that . 3loglog 2 =+ xy yx

246


Problem 8: Find if 66log44 a=3log2 , b=11log3 .

Problem 9: Simplify 21

10110 10loglog

101loglog 2

−⋅⋅⋅ aa aa .

Problem 10: (1989 NEAML) Determine the numerical value of: ( )( ))(log)(log 45

2 xy yx.

Problem 11: Solve . 10loglog 82 =+ xx Problem 12: Solve 6logloglog

3133 =++ xxx .

Problem 13: Solve

⎪⎩

⎪⎨

⎧

=++=++=++

(3). 2logloglog(2) 2logloglog(1) 2logloglog

16164

993

442

yxzxzyzyx

Problem 14: Solve . 5)312(log2 =−+ xx Problem 15: a, b, and c (c is the hypotenuse) are the lengths of three sides of a right triangle. Show that aaaa bcbcbcbc )()()()( loglog2loglog −+−+ ⋅=+ a ≠ 1.

Problem 16: Find if 45log4 a=10log3 and b=25log6 .

Problem 17: Find if ab ba loglog −3

10loglog =+ ab ba with a > b > 1.

247


Problem 18: For positive numbers x and y, if xy = 490 and Problem 19: Compare 2x, 3y, and 5z if . x, y, z are positive numbers. zyx 532 == Problem 20: Find the greatest value of log x + log y if 2x + 5y = 20. Problem 21: Find the smallest value of (log x)2 + (log y)2 if xy2 = 100 with 1 ≤ x ≤ 10. Problem 22: (1989 AIME) Find if 2

2 )(log x ).(loglog)(loglog 2882 xx = Problem 23: (2000 AIME I) The system of equations 4))(log(log)2000(log 101010 =− yxxy

1))(log(log)2(log 101010 =− zyyz

0))(log(log)(log 101010 =− xzzx

has two solutions ( ) and ( ). Find 111 ,, zyx 222 ,, zyx .21 yy +

248


SOLUTIONS TO PROBLEMS Problem 1: Solution: (C). Method 1: The first three given logarithmic equalities are equivalent to the following exponential equalities a = xp, b = xq, c = xr. Hence

yrpqrp

q

xxxx

acb

=== −−+

222

, y = 2q – p – r.

Method 2: We may express the relationship involving y in logarithmic form:

.logloglog2log cabxy −−= Substituting the given first relations for the logarithms on the right hand side of the equation yields y log x = 2q log x − p log x − r log x. Since x ≠ 1 and log x ≠ 0, division by log x yields y = 2q – p – r. Problem 2: Solution: (B). The identity together with the given equation yields

1))(log(log =NM MN

.12 =)(log MN

∴ 1 or – 1. If log =MN 1log =MN

.1−=

, then M = N, which is ruled out, so we may

conclude that log MN

∴ , MN = 1. 1−= NM Problem 3: Solution: 0.

429384 32log)2log2)(log3log3(log −++

45

23333

2log)2log212)(log

2log31

2log21( −++= = 0

45

61

41

31

21

=−+++ .

Problem 4: Solution: 5 and 25. We may rewrite 325loglog2 25 =+ xx as

249


3log

1log225

25 =+x

x ⇒ ⇒

01log3log2 25225 =+− xx

0)1)(log1log2( 2525 =−− xx

So ,21log25 =x 5252

1

1 ==x ; ,1log25 =x 252 =x .

We can check to see that these two values are both solutions by plugging them into the given equation. Problem 5: Solution: (C). Since for any base b ≠ 0, 0log =Nb only if N = 1, the given equations yield

.1)(loglog)(loglog)(loglog 322443 === zyx

,3log4 =x ,4log2

Moreover, since , only if

M = b, we have

1log =Mb

=y ,2log3 =z or equivalently x = 43, y = 24, z = 32. Adding these results gives x + y + z = 43 + 24 + 32 = 64 + 16 + 9 = 89. Problem 6: Solution: 9.

28log

log4log8log

3log4log

=⋅⋅m ⇒ 9log3log2log ==m ⇒ m = 9.

Problem 7: Solution: 108.

3loglog 2 =+ xy yx ⇒ 3log2log =+ xy yx ⇒ 3log

2log =+y

yx

x ⇒

⇒ ⇒ .

y (log x

0)2 =yx )(log 2 +

1(log −yx

xlog32 =)(log −yx

02log3)2 =+− yy x

Thus or 01log =−yx 02log =−yx . The solutions to the first equation are the 99 ordered pairs from (2, 2) to (100, 100); the solutions to the second equation are the 9 ordered pairs (2, 4), (3, 9), (4, 16),…,(10, 100). Thus there are 99 + 9 = 108 ordered pairs of solutions.

Problem 8: Solution: 1+ a + ab

2 + ab.

We have 11log2log211log12log

11log2log11log3log2log

44log66log66log

33

33

32

3

333

3

344 +

++=

+++

==

250


and ⇒ a=3log2 a13log2 =

Therefore ab

aba

ba

ba

+++

=+⋅

++=

21

12

11

66log44 .

Problem 9: Solution: – ½.

21010 loglog2log aaa ==

22

log110log

101log 2 aaa −=−=

aaa logloglog1)

101(

101 −=−=

−

aaa log2110log

2110log 2

1

−=−=−

.

Therefore

21

101

210 10loglog

101loglog

−⋅⋅⋅ aa aa

21)

log21)(log)(

log1(log 2

2 −=−−−=a

aa

a .

Problem 10: Solution (official solution): ( )( ))(log)(log 45

2 xy yx

= ( )(log)(log21 45 xy yx ⎟

⎠⎞

⎜⎝⎛ ) (Step 1)

= ( )( xy yx loglog)45(21

⋅⎟⎠⎞

⎜⎝⎛ ) (Step 2)

10loglog

loglog10 =⎟⎟

⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛=

yx

xy .

Note: We believe that step 1 is incorrect. In the given expression ( )( ))(log)(log 45

2 xy yx, x

can be either positive or negative. In step 1, x was forced to be positive in ⎟⎠⎞

⎜⎝⎛ )(log

21 5yx .

However, it is okay set ( ) )(log5)(log 225 yy xx = because y must be positive since y is the

base in the expression ( ))(log 4xy .

251


Problem 11: Solution: Method 1 (incorrect way): By Law 3: the Power Identity, loga xr

= r loga x, the given equation can be written as ⇒ 10 log x = 10 ⇒ log x = 1. 10log8log2 =+ xxTherefore, the solution is x = 10. We can plug this back into the given equation to check and see that it is a solution. However, we missed one solution because we misused the law 3, since we are not allowed to move “r” to the front if x > 0. Method 2 (correct way): We know that x2 and x8 are positive, so by Law 1: the Product Identity log a x + log a y

= loga xy , the given equation becomes log( ⇒ x10 = 1010 ⇒ x = ± 10. 10)82 =⋅ xxPlugging these values back into the given equation, we can check and see that both values of x are the solutions. Problem 12: Solution: 27.

We know that bmnb a

mam loglog = .

∴ ,loglog 233 xx = 1

3331 logloglog 1

−== − xxx .

The given equation becomes ⇒ 6logloglog 13

233 =++ −xxx 6log 2

3 =x

⇒ x2 = 36 ⇒ x = ± 27 . x = – 27 is extraneous, so the value for x is x = 27.

Problem 13: Solution: 32

=x , 827

=y , 3

32=z .

We know that x > 0, y > 0, and z > 0. Use the Change-of-Base formula, we can change the bases to 2, 3, and 4 in (1), (2), and (3), respectively:

252


⎪⎪⎪

⎩

⎪⎪⎪

⎨

⎧

=++

=++

=++

16.loglog21log

21log

9loglog21log

21log

4,loglog21log

21log

4444

3333

2222

yxz

xzy

zyx

⇒

⎪⎪⎩

⎪⎪⎨

⎧

=

=

=

(6). 16

(5), 9

(4), 4

xyz

zxy

yzx

Multiplication of all three equations together gives (xyz)2 = 242. Since x > 0, y > 0, z > 0, xyz = 24 (7) Square both sides of (4): x2yz = 16 (8)

Dividing equation (8) by equation (7), we get 32

=x .

Similarly, we can obtain the values of y and z:827

=y and 3

32=z .

Plugging these values back into the given equation, we can check and see that the values

32

=x , 827

=y , 3

32=z are the solutions.

Problem 14: Solution: 5. The given equation can be written as or 5)312(log2log 22 =−+ xx 5)]312(2[log2 =−xx

Therefore . Factoring, we get (2x + 1)(2x – 32) = 0. 52 2)2(31)2( =− xx

Since 2x + 1 > 0, we have 2x – 32 = 0. The answer is x = 5. Problem 15: Solution: Because a, b, and c are the lengths of three sides of a right triangle and c is the hypotenuse, rearranging the terms in the Pythagorean Theorem, we get

))((222 bcbcbca −+=−= .

)(log1

)(log1loglog )()( bcbc

aaaa

bcbc −+

+=+ −+

aa

bcbcbcbcbcbc

bcbc

a

aa

aa

)()( log1

log1

))((log)(log)(log)(log)(log

−+

⋅

−+=

−+++−

=

aaaaa bcbcbcbca )()()()(2 loglog2logloglog −+−+ ⋅=⋅⋅= .

253


Problem 16: Solution: ab + 3b + 42(2a − b)

.

)1( 2log25log245log

3

34

+=

We know that . 5log2log10log 333 +=

∴ (2) 5log2log 33 a=+

We also know that 2log1

5log225log3

36 +

= . ∴ (3) 2log1

5log2

3

3 b=+

Solving (2) and (3) for , we get ,5log3 2log3

(4) 2

22log3 +−

=b

ba

(5) 2

5log3 ++

=b

bab

Substituting (4) and (5) into (1) yields .)2(24345log4 ba

bab−

++=

Problem 17: Solution: −83

.

Method 1: ( )2 = ab ba loglog − abab baba loglog4)log(log 2 ⋅−+

9644)

310( 2 =−= .

We know that a > b > 1, so ab ba log1log << ⇒ 0loglog <− ab ba .

38loglog −=− ab ba .

Method 2: Since a > b > 1, . 1log >ab

We also know that 3

10loglog =+ ab ba ⇒ 313

log1log +=+

aa

bb .

∴ ,3log =ab ,31log =ba 3

8331loglog −=−=− ab ba .

254


Problem 18: Solution: 8. xy = 490, x > 0, y > 0 (1)

255

1∴ 7log2loglog +=+ yx

1)7log(log)7log(log =−+− yx (2)

We also have 4

143)7log(log)7log(log −=−+− yx (3)

From (2) and (3), we know that 7loglog −x and 7loglog −y are the two roots of

04

1432 =−− tt .

0)2

13)(2

11( =−+ tt .

Let x > y, 2

137loglog =−= xt .

∴ 5.67log2

137loglog +=+=x (4)

Since 1071021

<< , then .17log5.0 << From (4) we have 5.7log7 << x , so the integer part is 8. Problem 19: Solution: 5z > 2x > 3y. Since , we have zyx 532 == .5log3log2log kzyx ===

We know that x, y, z > 0, ∴ k > 0.

We have ,2log

kx = ,3log

kx = 5log

kz = .

∴ 03log2log8log9log32 >

−=− kyx ⇒ 2x > 3y (1)

05log2log32log25log52 <

−=− kzx ⇒ 2x < 5z (2)

Therefore 5z > 2x > 3y.


Problem 20: Solution: 1. We know that x > 0, y > 0, so log x + log y = log (xy). log x + log y will be the greatest if xy is the greatest. From AM-GM inequality, we have

xyyxyx 1025225220 =⋅≥+= Simplifying this inequality gives us xy ≤ 10, and so the greatest value of xy is 10 (when x = 5 and y = 2). Thus, the greatest value of log (xy) is 1 and the greatest value of log x + log y is 1.

Problem 21: Solution: 54 .

We know that y > 0 and x

y 10= .

222

222 )log211()(log10log)(log)(log)(log xx

xxyx −+=⎟

⎠

⎞⎜⎝

⎛+=+

54

52log

451log)(log

45 2

2 +⎟⎠⎞

⎜⎝⎛ −=+−= xxx

Since 1 ≤ x ≤ 10, then 0 ≤ log x ≤ 1.

When 52log =x , or 5

2

10=x , (log x)2 + (log y)2 has the smallest value of 54 .

Problem 22: Solution: 27.

Since xxxx

28 log31

2log31

8log1log === ,

)(loglog31)(loglog 2228 xx = (1)

Let . xy 2log=

(1) becomes: yy22 log

31

3log = ⇒ yy

23

2 log)3

(log = ⇒ yy=3)

3(

Rearranging the terms, we get . Since 0)27( 2 =−yy 0≠y , . 27)(log 22

2 == xy

256


257

z10

,2,2

,2,2

20

Problem 23: Solution: 25. Let u , , . x10 v log=log= y10 w log=

We rewrite the given system of equations as

or ⎪⎩

⎪⎨

⎧

=+−−=+−−

=+−−

.11log1

log1

10

10

uwwuwvvwvuuv

⎪⎩

⎪⎨

⎧

=−−=−−=−−

.1)1)(1(log)1)(1(log)1)(1(

10

10

uwwvvu

From the first two equations, we get u = w. From the third equation we have two cases: u = v = 2 or u = v = 0. Case I: . This gives us log10=v ,1001 =x 201 =y , 1001 =z .

Case II: . This gives us 510log=v ,12 =x 52 =y , 12 =z . Therefore y1 + y2 = 25.

50 amc lectures chapter 33 logarithms basic … amc lectures chapter 33 logarithms basic knowledge....

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