Question

part 1 of 2
use the change-of-base formula to approximate the logarithms to four decimal places.
\log_{7} 13 \approx 1.3181
part: 1 / 2
part 2 of 2
check your answer by using the exponential key on your calculator.
\square \approx 13

Explanation:

Step1: Recall the definition of logarithm

By the definition of logarithm, if $\log_{b}a = x$, then $b^{x}=a$. Here, we have $\log_{7}13\approx1.3181$, so we need to check if $7^{1.3181}\approx13$.

Step2: Use the exponential key on calculator

Using a calculator, we calculate $7^{1.3181}$. Let's compute this value. First, we know that $7^{1}=7$, $7^{1.3181}$ can be calculated as $e^{1.3181\times\ln(7)}$ (using the identity $a^{x}=e^{x\ln(a)}$). $\ln(7)\approx1.9459$, so $1.3181\times1.9459\approx2.5649$. Then $e^{2.5649}\approx13$ (since $e^{2.5649}\approx13$). So we fill in the base as 7 and the exponent as 1.3181 to check, i.e., $7^{1.3181}\approx13$.

Answer:

We need to fill in $7$ in the box (the base of the logarithm) and the exponent as $1.3181$, so the expression is $7^{1.3181}\approx13$. So the number in the box (the base) is $7$.