How do you solve for x in $log_2(x + 1) = 3$ ?

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Answer 1 · 2016-01-25T19:52:55

$x=7$

For the general case:
$color(white)("XX")color(red)(log_b a = c) hArr color(blue)(b^c=a)$

Therefore
$color(white)("XX")color(red)(log_2(x+1) = 3) hArr color(blue)(2^3 = x+1)$

$color(white)("XX")rarr x+1 = 8$

$color(white)("XX")rarr x=7$

How do you solve for x in log_2(x + 1) = 3log_2(x + 1) = 3?