How do you solve $25^{x} = 5^{x^{2} - 15}$ $25^x = 5^(x^2 - 15)$ ?

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Answer 1 · 2016-03-28T11:55:02

$x = {- 3, 5}$ $x={-3,5}$

$25^{x} = 5^{x^{2} - 15}$ $25^x=5^(x^2-15)$

$5^{2 x} = 5^{x^{2} - 15}$ $color(red)(5)^(2x)=color(red)(5)^(x^2-15)$
$equation has the same base value$ $"equation has the same base value"$

$2 x = x^{2} - 15$ $2x=x^2-15$

$x^{2} - 2 x - 15 = 0$ $x^2-2x-15=0$

$(x - 5) (x + 3) = 0$ $(x-5)(x+3)=0$
$if (x-5)=0" then " x=5$
$if(x+3)=0" then "x=-3$
$x={-3,5}$

How do you solve 25^x = 5^(x^2 - 15)25x=5x2−1525^x = 5^(x^2 - 15)?