How do you solve $16 = 2^{x + 1}$ $16=2^(x+1)$ ?

1 Answer

Oct 3, 2015

I found $x = 3$ $x=3$

Explanation:

Write it as:
$2^{4} = 2^{x + 1}$ $2^4=2^(x+1)$
take ${log}_{2}$ $log_2$ on both sides to get rid of the $2$ $2$ :
$4 = x + 1$ $4=x+1$
$x = 4 - 1 = 3$ $x=4-1=3$

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How do you solve $16 = 2^{x + 1}$ $16=2^(x+1)$ ?

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