Solve the exponential equation algebraically. Approximate the result to three deci- mal places. $$e ^ { x } - 7 = 16$$
A) $\ln 23 \approx 3.135$
B) $\ln 23 \approx - 3.135$
C) $\ln 23 \approx - 2.773$
D) $\ln 23 \approx 2.773$
E) $\ln 23 \approx 1.946$

Question

Solve the exponential equation algebraically. Approximate the result to three deci- mal places. $$e ^ { x } - 7 = 16$$ 
A) $\ln 23 \approx 3.135$
B) $\ln 23 \approx - 3.135$
C) $\ln 23 \approx - 2.773$
D) $\ln 23 \approx 2.773$
E) $\ln 23 \approx 1.946$

Quizplus · Accepted Answer

First, add 7 to both sides to get $e^x = 23$. Then, take the natural logarithm of both sides to solve for $x$, giving $x = \ln(23)$. Approximating $\ln(23)$ gives approximately 3.135.

Solve the Exponential Equation Algebraically ex−7=16e ^ { x } - 7 = 16ex−7=16

Solve the Exponential Equation Algebraically $e ^ { x } - 7 = 16$