How do you solve $5^{n - 1} = \frac{1}{25}$ $5^(n-1)=1/25$ ?

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Answer 1 · 2015-08-17T06:41:55

$n = - 1$ $color(blue)(n=-1$

$5^{n - 1} = \frac{1}{25}$ $5^(n-1)=1/25$

$\frac{1}{25} = \frac{1}{5^{2}}$ $1/25=1/5^2$

As per property
$\frac{1}{a} = a^{- 1}$ $color(blue)(1/a=a^-1$

So, $\frac{1}{5^{2}} = 5^{- 2}$ $1/5^2 = 5^-2$

$5^{n - 1} = \frac{1}{25}$ $5^(n-1)=1/25$

$5^{n - 1} = 5^{- 2}$ $5^(n-1)=5^-2$
Since the bases are same we equate the exponents

$n - 1 = - 2$ $n-1=-2$

$n = - 1$ $color(blue)(n=-1$

How do you solve 5^(n-1)=1/255n−1=125 5^(n-1)=1/25?