solving exponential equations equations with variables in exponents, such as 3 x = 5 and 7 3x = 90...

Solving Exponential Equations

Equations with variables in exponents, such as

3x = 5 and 73x = 90

are called exponential equations.

In Section 9.3, we solved certain logarithmic equations by using the principle

am = x means logax = m

Solution

Example Solve: 3 x +1 = 43

We have 3 x +1 = 43

log 3 x +1 = log 43

(x +1)log 3 = log 43

x 2.424.

Principle of logarithmic equality

Power rule for logs

log 431

log3x

log 431

log3x Shuhaw Answer

Chemistry Answer

Solve graphically

x 2.424

Solution

Example Solve: e1.32t = 2000

We have:

t 5.758.

Note that we use the natural logarithm

Logarithmic and exponential functions are inverses of each other

e1.32t = 2000

ln e1.32t = ln 2000

1.32t = ln 2000

ln 2000

1.32t

To Solve an Equation of the Form at = b for t1. Take the logarithm (either natural or common) of both sides.

2. Use the power rule for exponents so that the variable is no longer written as an exponent.

3. Divide both sides by the coefficient of the variable to isolate the variable.

4. If appropriate, use a calculator to find an approximate solution in decimal form.

ln lnta b log logta b

ln lnt a b log logt a b

ln

ln

bt

a

log

log

bt

a

Solution

Example Solve: log2(6x + 5) = 4.

6x + 5 = 24

6x = 11

The solution is x = 11/6.

log2(6x + 5) = 4

6x + 5 = 16

x = 11/6

logax = m means am = x

Solving Logarithmic Equations

Solve graphically

x = 11/6

Using change of base 1ln(6 5)

(4)ln(2)

xY

Solution

Example Solve: log x + log (x + 9) = 1.

x2 + 9x = 10

To increase the understanding, we write in the base 10.

log10 x + log10 (x + 9) = 1

log10[x(x + 9)] = 1

x(x + 9) = 101

x2 + 9x – 10 = 0

(x – 1)(x + 10) = 0

x – 1 = 0 or x + 10 = 0 x = 1 or x = –10

0 + log (10) = 1

x = 1:

log 1 + log (1 + 9) = 1

0 + 1 = 1 TRUE

x = –10:

log (–10) + log (–10 + 9) = 1 FALSE

The solution is x = 1.

The logarithm of a negative number is undefined.

Solve graphically

We graph y1= log(x) + log (x + 9) - (1)

x = 1

Solution

Example Solve: log3(2x + 3) – log3(x – 1) = 2.

log3(2x + 3) – log3(x – 1) = 2

(2x + 3) = 9(x – 1)

x = 12/7

2x + 3 = 9x – 9

The solution is 12/7. Check is left to the student.

32 3

log 21

x

x

22 33

1

x

x

2 39

1

x

x

Solution