Solve: log₉(e³) × (log₂24 - log₂8) × ln(8×2) | Complete Logarithm Problem

$\log_9e^3\times(\log_224-\log_28)(\ln8+\ln2)$

We will solve the problem step by step:

Step 1: Simplify $\log_9 e^3$

• Using the change of base formula, $\log_9 e^3 = \frac{\ln e^3}{\ln 9}$.
• We know $\ln e^3 = 3\ln e = 3$, because $\ln e = 1$.
• Thus, $\log_9 e^3 = \frac{3}{\ln 9} = \frac{3}{2\ln 3}$, since $\ln 9 = 2\ln 3$.
• Therefore, $\log_9 e^3 = \frac{3}{2\ln 3}$.

Step 2: Simplify $\log_2 24 - \log_2 8$

• Use the logarithm subtraction rule: $\log_2 24 - \log_2 8 = \log_2 \left(\frac{24}{8}\right) = \log_2 3$.

Step 3: Simplify $\ln 8 + \ln 2$

• Using the product property of logarithms: $\ln 8 + \ln 2 = \ln(8 \times 2) = \ln 16$.
• Since $16 = 2^4$, $\ln 16 = 4\ln 2$.

Step 4: Combine the results

• We need to check the overall structure: $\log_9 e^3 \times \log_2 3 \times 4 \ln 2$.
• Previously calculated: $\log_9 e^3 = \frac{3}{2 \ln 3}$, $\log_2 3 = \frac{\ln 3}{\ln 2}$.
• Therefore, the entire expression becomes:
• $\frac{3}{2 \ln 3} \times \frac{\ln 3}{\ln 2} \times 4 \ln 2 = \frac{3}{2} \times 4 = 6$.

Therefore, the solution to the problem is $6$.