Question

Evaluate log7 49^2 + log7 7

Answer 1

To evaluate \( \log_7 49^2 + \log_7 7 \), we can use the properties of logarithms.

First, recall that \( 49 \) can be expressed as \( 7^2 \):

\[
49 = 7^2
\]

Next, we replace \( 49^2 \):

\[
49^2 = (7^2)^2 = 7^{2 \cdot 2} = 7^4
\]

Now, we can substitute this back into our expression:

\[
\log_7 49^2 = \log_7 (7^4)
\]

Using the property of logarithms that states \( \log_b (b^x) = x \), we have:

\[
\log_7 (7^4) = 4
\]

Now, for the second term \( \log_7 7 \):

\[
\log_7 7 = 1
\]

Putting it all together, we can now evaluate:

\[
\log_7 49^2 + \log_7 7 = 4 + 1 = 5
\]

Thus, the final answer is:

\[
\boxed{5}
\]