How do you solve $16^{5 x} = 64^{x + 7}$ $16^(5x)= 64^(x+7)$ ?

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Answer 1 · 2015-07-19T13:47:44

$x = 3$ $color(green)(x=3$

expressing the bases in terms of $4$ $4$

$16 = 4^{2}$ $16=4^2$

$64 = 4^{3}$ $64=4^3$

Rewriting the expression:
$16^{5 x} = 64^{x + 7}$ $16^(5x)=64^(x+7)$

$4^{2 \cdot (5 x)} = 4^{3 \cdot (x + 7)}$ $4^(2*(5x))=4^(3*(x+7))$

$4^{10 x} = 4^{3 x + 21}$ $4^(10x)=4^(3x+21)$

Now we can equate the powers:

$10 x = 3 x + 21$ $10x=3x+21$

$7 x = 21$ $7x=21$

$x = 3$ $color(green)(x=3$

How do you solve 165x=64x+716^(5x)= 64^(x+7)?