# Precalculus

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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 -

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 -

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 -

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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