Question

f(x)=log5 x

Evaluate f (1/25)

Answer 1

To evaluate \( f\left(\frac{1}{25}\right) \) for the function \( f(x) = \log_5 x \), we can substitute \( \frac{1}{25} \) into the function:

\[
f\left(\frac{1}{25}\right) = \log_5\left(\frac{1}{25}\right)
\]

We can rewrite \( \frac{1}{25} \) as \( 25^{-1} \). Since \( 25 = 5^2 \), we have:

\[
\frac{1}{25} = (5^2)^{-1} = 5^{-2}
\]

Now we can use the logarithmic property that states \( \log_b(a^c) = c \cdot \log_b(a) \):

\[
\log_5\left(\frac{1}{25}\right) = \log_5(5^{-2}) = -2 \cdot \log_5(5)
\]

Since \( \log_5(5) = 1 \), we find:

\[
-2 \cdot \log_5(5) = -2 \cdot 1 = -2
\]

Thus, the value of \( f\left(\frac{1}{25}\right) \) is:

\[
\boxed{-2}
\]