College Algebra
posted by Hope .
How do you solve 3 log(base 2)*(x1)+ log(base 2)4=5

recall that n*log(a) = log(a^n)
so, using base 2, we have
3log(x1)+log(4) = 5
log(x1)^3 + 2 = 5
log(x1)^3 = 3
(x1)^3 = 2^3
x1 = 2
x = 3
check:
3log2 + log4 = 5
3+2 = 5 
another way would be
3log(x1)+log(4) = 5
3log(x1) + 2 = 5
3log(x1) = 3
log(x1) = 1
x1 = 2^1 = 2
x = 3