How do you solve $7^(x– 5) = 7^(4x + 7)$ ?

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Answer 1 · 2015-07-25T14:34:45

$color(blue)(x=-4$

$7^(x–5)=7^(4x+7)$

Here, as the bases are common we can equate the powers to obtain $x$

$x–5=4x+7$

$–5-7=4x -x$

$3x=-12$

$color(blue)(x=-4$

How do you solve 7^(x– 5) = 7^(4x + 7)7^(x– 5) = 7^(4x + 7)?