Example

Example 10.25: Solving log2(3x5)=4\log_2(3x - 5) = 4 and lne2x=4\ln e^{2x} = 4

To solve the logarithmic equation log2(3x5)=4\log_2(3x - 5) = 4, first rewrite it in its equivalent exponential form, 24=3x52^4 = 3x - 5. Simplifying the exponential expression gives 16=3x516 = 3x - 5. Adding 55 to both sides results in 21=3x21 = 3x, and dividing by 33 yields x=7x = 7. To verify the solution, substitute x=7x = 7 back into the original equation: log2(375)=log2(215)=log2(16)\log_2(3 \cdot 7 - 5) = \log_2(21 - 5) = \log_2(16), which evaluates to 44, confirming the solution is correct.

To solve the natural logarithmic equation lne2x=4\ln e^{2x} = 4, rewrite it in exponential form as e4=e2xe^4 = e^{2x}. Because the bases on both sides of the equation are identical, the exponents must be equal, leading to 4=2x4 = 2x. Dividing by 22 yields x=2x = 2. Substituting x=2x = 2 back into the original equation gives lne22=lne4\ln e^{2 \cdot 2} = \ln e^4, which simplifies to 44, verifying the solution.

0

1

Updated 2026-05-25

Contributors are:

Who are from:

Tags

OpenStax

Intermediate Algebra @ OpenStax

Ch.10 Exponential and Logarithmic Functions - Intermediate Algebra @ OpenStax

Algebra

Related