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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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