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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.
Solving by Equating Exponents
SOLUTION
3x 3x x +x1+ 1
4
Solve 4= 8= 8
.
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.
Power of a power property
22 (3x) = 23(x + 1)
3•1
1+1
4
=
8
Solve for x.
6x
3x + 3
2 =2
Solution checks.
64 = 64
Equate exponents.
6x = 3x + 3
x =1
The solution is 1.
Solve for x.
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.
Taking a Logarithm of Each Side
S
OLUTION
Solve
10 2x – 3 + 4 = 21.
10 2x – 3 + 4 = 21
10 2x – 3 = 17
log 10 2x – 3 = log 17
2x – 3 = log 17
2x = 3 + log 17
Write original equation.
Subtract 4 from each side.
Take common log of each side.
log 10 x = x
Add 3 to each side.
x = 1 (3 + log 17)
2
Multiply each side by 1 .
x  2.115
Use a calculator.
2
Taking a Logarithm of Each Side
Solve 10 2x – 3 + 4 = 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  2.115.
y
y = 10 2x – 3 + 4
y = 21
1.0
2.0
x
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.
Solving a Logarithmic Equation
Solve log 3 (5x – 1) = log 3 (x + 7) .
SOLUTION
CHECK
Check the solution by substituting it into the original equation.
log 3 (5x – 1) = log 3 (x + 7)
log 3 (5x – 1) = log 3 (x + 7)
5x – 1 = x + 7
?
log 3 (5 · 2 – 1) = log 3 (2 + 7)
5x = x + 8
log 3 9 = log 3 9
x=2
The solution is 2.
Write original equation.
Write original equation.
Use property for logarithms
with the same base.
Substitute 2 for x.
Add 1 to each side.
Solution checks.
Solve for x.
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
log 5 (3x + 1) = 2
log5 (3x + 1)
?2
5
=
5
log 5 (3 · 8 + 1) = 2
?
3x
= 25
log+ 15 25
=2
x =28= 2
The solution is 8.
Write original equation.
Write original equation.
Exponentiate each side using base 5.
Substitute 8 for x.
b logSimplify.
bx = x
Solution
Solve
for x. checks.
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.
Checking for Extraneous Solutions
SOLUTION
Check for
extraneous
equation.
log
5xlog
+ log
– 1)(x= –2 1) = 2Write
Solve
5x (x
+ log
. original
solutions.
log [ 5x (x – 1)] = 2
10
log (5x 2 – 5x)
Product property of logarithms.
= 10 2 Exponentiate each side using
base 10.
5x 2 – 5x = 100
x 2 – x – 20 = 0
(x – 5)(x + 4) = 0
x = 5 or x = –4
10 log x = x
Write in standard form.
Factor.
Zero product property
Checking for Extraneous Solutions
SOLUTION
log 5x + log (x – 1) = 2
x = 5 or x = –4
Check for extraneous solutions.
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
y=2
x
y = log 5x + log (x – 1)
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 = 0.291 ln E + 1.17
Using a Logarithmic Model
SOLUTION
M = 0.291 ln E + 1.17 Write model for moment magnitude.
9.5 = 0.291 ln E + 1.17 Substitute 9.5 for M.
8.33 = 0.291 ln E
Subtract 1.17 from each side.
28.625  ln E
Divide each side by 0.291.
e 28.625  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.
SOLVING LOGARITHMIC EQUATIONS
EXPONENTIAL AND LOGARITHMIC PROPERTIES
x
y
1
For b > 0 and b  1, if b = b , then x = y.
2
For positive numbers b, x, and y where b  1,
log b x = log b y if and only if x = y.
3
x
y
For b > 0 and b  1, if x = y, then b = b .