Question

solve the following equation for x:
$e^{3x - 1}=4$
a. 1
b. $\frac{1 + 4e}{3}$
c. $\frac{ln 5}{3}$
d. $\frac{1+ln 4}{3}$

Explanation:

Step1: Take natural logarithm on both sides

$$\ln(e^{3x - 1})=\ln(4)$$
Using the property $\ln(e^{a})=a$, we get $3x - 1=\ln(4)$.

Step2: Solve for \(x\)

Add \(1\) to both sides: $3x=\ln(4)+1$.
Divide both sides by \(3\): $x = \frac{1+\ln(4)}{3}$.

Answer:

D. \(\frac{1+\ln4}{3}\)