Asked by Hope
How do you solve 3 log(base 2)*(x-1)+ log(base 2)4=5
Answers
Answered by
Steve
recall 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
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
Answered by
Steve
another 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
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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