Precalculus
Exponential and Logarithmic Functions
Solving Exponential and Logarithmic Equations
Solving Exponential Equations
Concepts 1
Example 1.
Solve
3
2x-5
= 27
.
Solution
By observing that
27 = 3
3
, we have the same base on both sides of the equation:
3
2x-5
= 3
3
Since an exponential function is a one-to-one function, we can equate exponents:
2x-5 = 3
2x = 3 + 5
2x = 8
Answer: x = 4.
Example 2.
Solve
2
2x
- 5·2
x
- 24 = 0
.
Solution
The given equation can be interpreted as a quadratic equation in 2
x
. Using the quadratic formula to solve for 2
x
yields
We conclude that either 2
x
= 8 or 2
x
= -3. However, 2
x
= -3 must be ruled out since 2
x
is always positive.
2
x
= 8
2
x
= 2
3
x = 3
Answer: x = 3.
Example 3.
Solve
2
x
= 9
.
Use a calculator to find x to the nearest 0.0001.
Solution
Taking the base 10 logarithm of both sides of the given equation, we see that
log
10
2
x
= log
10
9
x·log
10
2 = log
10
9
Answer: x = 3.1699.
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