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Answer :
We start with the logarithmic equation:
[tex]$$
\log_5 \frac{1}{25} = -2.
$$[/tex]
Recall the definition of logarithms: if
[tex]$$
\log_b (a) = c,
$$[/tex]
then the equivalent exponential form is
[tex]$$
a = b^c.
$$[/tex]
Using this definition with [tex]$a = \frac{1}{25}$[/tex], [tex]$b = 5$[/tex], and [tex]$c = -2$[/tex], we rewrite the equation as:
[tex]$$
\frac{1}{25} = 5^{-2}.
$$[/tex]
This is the logarithmic equation in exponential form.
To check, we note that:
[tex]$$
5^{-2} = \frac{1}{5^2} = \frac{1}{25},
$$[/tex]
which confirms that the conversion is correct.
[tex]$$
\log_5 \frac{1}{25} = -2.
$$[/tex]
Recall the definition of logarithms: if
[tex]$$
\log_b (a) = c,
$$[/tex]
then the equivalent exponential form is
[tex]$$
a = b^c.
$$[/tex]
Using this definition with [tex]$a = \frac{1}{25}$[/tex], [tex]$b = 5$[/tex], and [tex]$c = -2$[/tex], we rewrite the equation as:
[tex]$$
\frac{1}{25} = 5^{-2}.
$$[/tex]
This is the logarithmic equation in exponential form.
To check, we note that:
[tex]$$
5^{-2} = \frac{1}{5^2} = \frac{1}{25},
$$[/tex]
which confirms that the conversion is correct.
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