Precalculus
posted by Melinda .
Explain how to do these please.
1. solve x=log(base2)*1/8 by rewriting in exponential form.
2. Solve log(3x+1)=5
3. Solve log(base x)8=1/2

x = log2 (1/2)^3
x = 3 log2 (1/2)
x/3 = log 2(1/2) = log 2 (1)  log2(2)
but log2 (1) = 0 and log 2 (2) = 1
so
x/3 = 1
x = 3 
log(3x+1)=5
I get we mean log base 10
10^log(3x+1) = 3x+1
so
3x+1 = 10^5
3 x = 10^5 1 = 9999
x = 3333 
log(base x)8=1/2
x^log(base x)8 = 8
so
8 = x^(1/2) = 1/sqrt x
sqrt x = 1/8
x = 1/64