How do you solve $25^{x - 1} = 125^{4 x}$ $25^(x-1)=125^(4x)$ ?

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Answer 1 · 2016-08-19T05:39:44

$x = - \frac{1}{5}$ $x=-1/5$

Since $25 = 5^{2}$ $25=5^2$ and $125 = 5^{3}$ $125=5^3$ you can rewrite:

$5^{2 (x - 1)} = 5^{3 (4 x)}$ $5^(2(x-1))=5^(3(4x))$

that implies:

$2 (x - 1) = 3 (4 x)$ $2(x-1)=3(4x)$

that's

$2 x - 2 = 12 x$ $2x-2=12x$

$10 x = - 2$ $10x=-2$

$x = - \frac{1}{5}$ $x=-1/5$

How do you solve 25^(x-1)=125^(4x)25x−1=1254x25^(x-1)=125^(4x)?