Posted by **Melinda** on Sunday, January 29, 2012 at 12:45pm.

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

- Precalculus -
**Damon**, Sunday, January 29, 2012 at 1:43pm
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

- Precalculus -
**Damon**, Sunday, January 29, 2012 at 1:45pm
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

- Precalculus -
**Damon**, Sunday, January 29, 2012 at 1:47pm
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

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