How do you solve $25^{x} = 125^{x - 2}$ $25^x=125^(x-2)$ ?

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Answer 1 · 2017-06-05T07:15:03

$x = 6$ $x=6$

In this exponential equation, the clue is that both $25 and 125$ $25 and 125$ are powers of $5$ $5$ .

Write them both with a base of $5$ $5$

$25^{x} = 125^{x - 2}$ $25^x = 125^(x-2)$

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

Raising a power to a power $\to$ $rarr$ multiply the indices:

$5^{(2 x)} = 5^{(3 x - 6)}$ $color(blue)(5)^color(red)((2x)) = color(blue)(5)^color(red)((3x-6))$

In this equation the bases are both equal (to $5$ $5$ ,) so the indices also have to be equal.

$:. color(red)(2x=3x-6)" "larr$ solve this linear equation for $x$

$6=x$

How do you solve 25^x=125^(x-2)25x=125x−225^x=125^(x-2)?