How do you solve $2^{3 x + 19} = 16^{x + 8}$ $2^(3x+19) = 16^(x+8)$ ?

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Answer 1 · 2015-07-19T13:38:29

$x = - 13$ $color(blue)(x=-13$

$2^{3 x + 19} = 16^{x + 8}$ $2^(3x+19)=16^(x+8)$

$16 = 2^{4}$ $16= 2^4$
Rewriting the expression:

$2^{3 x + 19} = 2^{4 . (x + 8)}$ $2^(3x+19)=2^(4.(x+8)$

$2^{3 x + 19} = 2^{4 x + 32}$ $2^(3x+19)=2^(4x+32)$

Now we can equate the powers:

$3 x + 19 = 4 x + 32$ $3x+19 = 4x+32$

$x = = 19 - 32$ $x= = 19-32$
$x = - 13$ $color(blue)(x=-13$

How do you solve 2^(3x+19) = 16^(x+8)23x+19=16x+82^(3x+19) = 16^(x+8)?