Question

no calculator is allowed on this question. which of the following logarithmic expressions is equivalent to \\(\log_{2} e^{3x}\\)? select one answer a \\(3x + \log_{2} e\\) b \\(3x - \log_{2} e\\) c \\(3x \cdot \ln 2\\) d \\(\frac{3x}{\ln 2}\\)

Explanation:

Step1: Use logarithm power rule

The power rule of logarithms states that $\log_b a^n = n\log_b a$. For the expression $\log_2 e^{3x}$, we can apply this rule. So, $\log_2 e^{3x}=3x\log_2 e$.

Step2: Change of base formula

The change of base formula for logarithms is $\log_b a=\frac{\ln a}{\ln b}$. Applying this to $\log_2 e$, we get $\log_2 e = \frac{\ln e}{\ln 2}$. Since $\ln e = 1$, this simplifies to $\frac{1}{\ln 2}$.

Step3: Substitute back

Now substitute $\log_2 e=\frac{1}{\ln 2}$ into $3x\log_2 e$. We have $3x\times\frac{1}{\ln 2}=\frac{3x}{\ln 2}$.

Answer:

D. $\frac{3x}{\ln 2}$