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

1 Answer
Alan P.
Jan 25, 2016

#x=7#

Explanation:

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#

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