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December 22, 2014

December 22, 2014

Posted by **Hope** on Monday, November 12, 2012 at 12:15pm.

- College Algebra -
**Steve**, Monday, November 12, 2012 at 12:32pmrecall that n*log(a) = log(a^n)

so, using base 2, we have

3log(x-1)+log(4) = 5

log(x-1)^3 + 2 = 5

log(x-1)^3 = 3

(x-1)^3 = 2^3

x-1 = 2

x = 3

check:

3log2 + log4 = 5

3+2 = 5

- College Algebra -
**Steve**, Monday, November 12, 2012 at 12:34pmanother way would be

3log(x-1)+log(4) = 5

3log(x-1) + 2 = 5

3log(x-1) = 3

log(x-1) = 1

x-1 = 2^1 = 2

x = 3

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