Solve the exponential equation algebraically.Approximate the result to three decimal places. $4 e ^ { x } = 33$
A) $\ln \frac { 33 } { 4 } \approx 2.110$
B) $\ln \frac { 33 } { 4 } \approx - 2.110$
C) $\ln \frac { 4 } { 33 } \approx 2.110$
D) $\ln \frac { 33 } { 4 } \approx 3.497$
E) $\ln \frac { 33 } { 4 } \approx 0.114$

Question

Solve the exponential equation algebraically.Approximate the result to three decimal places.​ $4 e ^ { x } = 33$ ​
A) $\ln \frac { 33 } { 4 } \approx 2.110$
B) $\ln \frac { 33 } { 4 } \approx - 2.110$
C) $\ln \frac { 4 } { 33 } \approx 2.110$
D) $\ln \frac { 33 } { 4 } \approx 3.497$
E) $\ln \frac { 33 } { 4 } \approx 0.114$

Quizplus · Accepted Answer

To solve $4e^x = 33$, divide both sides by 4 to get $e^x = \frac{33}{4}$. Taking the natural logarithm of both sides gives $x = \ln\left(\frac{33}{4}\right)$. Calculating this value gives approximately 2.110.

Solve the Exponential Equation Algebraically 4ex=334 e ^ { x } = 334ex=33

Solve the Exponential Equation Algebraically $4 e ^ { x } = 33$