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To solve the equation \(7^{-8a+10} = 95\) by taking the logarithm of both sides, you'll want to use the property \(\log_b(x^y) = y \cdot \log_b(x)\). We'll take the logarithm of both sides with base 7.

Starting with the given equation:
\[7^{-8a+10} = 95\]

Taking the logarithm of both sides with base 7:
\[\log_7(7^{-8a+10}) = \log_7(95)\]

Using the property of logarithms:
\[-8a+10 = \log_7(95)\]

Now, to solve for \(a\), we isolate \(a\):
\[a = \frac{10 - \log_7(95)}{-8}\]

Now, we'll calculate the value of \(a\):

\[a = \frac{10 - \log_7(95)}{-8}\]

Using a calculator to evaluate \(\log_7(95)\), and then substituting it into the equation:

\[a \approx \frac{10 - 2.1271}{-8} \]

\[a \approx \frac{10 - 2.1271}{-8} \]

\[a \approx \frac{10 - 2.1271}{-8} \]

\[a \approx \frac{7.8729}{-8} \]

\[a \approx -0.9841\]

So, rounding to the nearest ten-thousandth, \(a \approx -0.9841\).
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