1. For b > 0 and b ≠ 1, if b x = b y, then x = y.
SOLVING EXPONENTIAL EQUATIONS
If two powers with the same base are equal, then their exponents must be equal.
For b > 0 and b ≠ 1, if b x = b y, then x = y.

2. Solve 43x = 8 x + 1. 4 3x = 8 x + 1 ( 22)3x = ( 23) x + 1
Solving by Equating Exponents
Solve 43x = 8 x + 1.
4 3x = 8 x + 1
SOLUTION
Write original equation.
( 22)3x = ( 23) x + 1
Rewrite each power with base 2 .
CHECK
Check the solution by substituting it into the original equation.
22 (3x) = 23(x + 1)
Power of a power property
4 3 • 1 =
Solve for x.
26x = 23x + 3
64 = 64
Solution checks.
6x = 3x + 3
Equate exponents.
x = 1
Solve for x.
The solution is 1.

3. Solving by Equating Exponents
When it is not convenient to write each side of an exponential equation using the same base, you can solve the equation by taking a logarithm of each side.

4. Taking a Logarithm of Each Side
Solve 10 2 x – = 21.
SOLUTION
10 2 x – = 21
Write original equation.
10 2 x – 3 = 17
Subtract 4 from each side.
log 10 2 x – 3 = log 17
Take common log of each side.
2 x – 3 = log 17
log 10 x = x
2 x = 3 + log 17
Add 3 to each side.
x = (3 + log 17 ) 1 2
Multiply each side by 1 2
x ≈ 2.115
Use a calculator.

5. Taking a Logarithm of Each Side
Solve 10 2 x – = 21.
CHECK
Check the solution algebraically by substituting into the original equation. Or, check it graphically by graphing both sides of the equation and observing that the two graphs intersect at x ≈
1.0
2.0 x y
y = 10 2 x – 3 + 4
y = 21

6. Log b x = log b y if and only if x = y.
SOLVING LOGARITHMIC EQUATIONS
To solve a logarithmic equation, use this property for logarithms with the same base:
For positive numbers b, x, and y where b ≠ 1,
Log b x = log b y if and only if x = y.

7. Use property for logarithms with the same base.
Solving a Logarithmic Equation
Solve log 3 (5 x – 1) = log 3 (x + 7) .
SOLUTION
CHECK
Check the solution by substituting it into the original equation.
log 3 (5 x – 1) = log 3 (x + 7)
Write original equation.
log 3 (5 x – 1) = log 3 (x + 7)
Write original equation.
5 x – 1 = x + 7
Use property for logarithms with the same base.
log 3 (5 · 2 – 1) = log 3 (2 + 7) ?
Substitute 2 for x.
5 x = x + 8
Add 1 to each side.
log 3 9 = log 3 9
Solution checks.
x = 2
Solve for x.
The solution is 2.

8. log5 (3x + 1) Solve log 5 (3x + 1) = 2 . log 5 (3x + 1) = 2
Solving a Logarithmic Equation
Solve log 5 (3x + 1) = 2 .
SOLUTION
CHECK
Check the solution by substituting it into the original equation.
log 5 (3x + 1) = 2
Write original equation.
log 5 (3x + 1) = 2
Write original equation.
log 5 (3 · 8 + 1) = 2 ?
= 5 2
log5 (3x + 1)
Exponentiate each side using base 5.
Substitute 8 for x.
log = 2 ?
3x + 1 = 25
b = x
log b x
Simplify.
2 = 2
x = 8
Solution checks.
Solve for x.
The solution is 8.

9. Checking for Extraneous Solutions
Because the domain of a logarithmic function generally does not include all real numbers, you
should be sure to check for extraneous solutions of logarithmic equations. You can do this algebraically
or graphically.

10. Check for extraneous solutions. log 5 x + log (x – 1) = 2
Checking for Extraneous Solutions
Check for extraneous solutions.
SOLUTION
log 5 x + log (x – 1) = 2
Solve log 5 x + log (x – 1) = 2 .
Write original equation.
log [ 5 x (x – 1)] = 2
Product property of logarithms.
= 10 2
log (5 x 2 – 5 x)
Exponentiate each side using base 10.
5 x 2 – 5 x = 100
10 log x = x
x 2 – x – 20 = 0
Write in standard form.
(x – 5 )(x + 4) = 0
Factor.
x = 5 or x = – 4
Zero product property

11. Check for extraneous solutions.
Checking for Extraneous Solutions
SOLUTION
log 5 x + log (x – 1) = 2
Check for extraneous solutions.
x = 5 or x = – 4
Zero product property
The solutions appear to be 5 and – 4. However, when you check these in the original equation or use a graphic check
as shown below, you can see that x = 5 is the only solution. y x
y = 2
y = log 5 x + log (x – 1)

12. Using a Logarithmic Model
SEISMOLOGY On May 22, 1960, a powerful earthquake took
place in Chile. It had a moment magnitude of 9.5.
How much energy did this earthquake release?
The moment magnitude M of an earthquake that releases energy E (in ergs) can be modeled by this equation:
M = ln E

13. The earthquake released about 2.7 trillion ergs of energy.
Using a Logarithmic Model
SOLUTION
M = ln E
Write model for moment magnitude.
9.5 = ln E
Substitute 9.5 for M.
8.33 = ln E
Subtract 1.17 from each side.
≈ ln E
Divide each side by 0.291
e ≈ e ln E
Exponentiate each side using base e.
2.702 x 1012 ≈ E
e ln x = e log e x = x
The earthquake released about 2.7 trillion ergs of energy.

14. SOLVING LOGARITHMIC EQUATIONS
EXPONENTIAL AND LOGARITHMIC PROPERTIES 1
For b > 0 and b ≠ 1, if b x = b y , then x = y.
For positive numbers b, x, and y where b ≠ 1,
log b x = log b y if and only if x = y. 2 3
For b > 0 and b ≠ 1, if x = y, then b x = b y .