How do you solve $2^{3 x + 5} = 128$ $2^(3x+5)=128$ ?

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Answer 1 · 2017-01-28T14:46:05

The solution is $x = \frac{2}{3}$ $x=2/3$

$128 = 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 = 2^{7}$ $128=2*2*2*2*2*2*2=2^7$

$2^{3 x + 5} = 2^{7}$ $2^(3x+5)=2^7$

So,

$3 x + 5 = 7$ $3x+5=7$

$3 x = 7 - 5 = 2$ $3x=7-5=2$

$x = \frac{2}{3}$ $x=2/3$

How do you solve 2^(3x+5)=12823x+5=1282^(3x+5)=128?