# If 5^-x = 1/3, then what is the value of 25^x?

Nov 26, 2017

$\approx 8.92$

#### Explanation:

First solve for x

1/5^× = 1/3
5^× = 3
$x \log \left(5\right) = \log \left(3\right)$
×=log(3)/log(5)~~.68
Therefore
25^× ~~8.92

So the only part I want to emphasize is the log operation. Log is a nice way to change multiplication and division into addition and subtraction. Another benefit is that any power becomes a multiple. Uaing it allowed me to bring x down and then solve more simply.

Nov 27, 2017

$9$

#### Explanation:

Alternate way of doing it.

We know that $25 = {5}^{2}$, thus ${25}^{x} = {\left({5}^{2}\right)}^{x} = {5}^{2 x} = {\left({5}^{x}\right)}^{2}$

We know from the first equation that $\frac{1}{5} ^ x = \frac{1}{3}$, because ${a}^{-} n = \frac{1}{a} ^ n$.

This means that ${5}^{x} = 3$. We rewrote ${25}^{x}$ as ${\left({5}^{x}\right)}^{2}$ above so now we can substitute.

${\left({5}^{x}\right)}^{2} = {3}^{2} = 9$

Hopefully this helps!