Precalculus Exponential and Logarithmic Functions Solving Exponential and Logarithmic Equations
Solving Exponential Equations Concepts 1

Example 1. Solve 32x-5 = 27 .

Solution
By observing that 27 = 33, we have the same base on both sides of the equation:
32x-5 = 33
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 22x - 5·2x - 24 = 0 .

Solution
The given equation can be interpreted as a quadratic equation in 2x. Using the quadratic formula to solve for 2x yields
We conclude that either 2x = 8 or 2x = -3. However, 2x = -3 must be ruled out since 2x is always positive.
2x = 8
2x = 23
x = 3

Answer: x = 3.

Example 3. Solve 2x = 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
log102x = log109
x·log102 = log109

Answer: x = 3.1699.

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