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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