Solve in terms of b: \log_b (1 - 3x) = 3

Question

Solve in terms of b:

\log_{b} (1 - 3 x) = 3 + \log_{b} x

Answer 1

Remember, any

\log

to the base b, such as

\log_{b} (y)

, is in its heart an exponent. Let us start, then, from your expression

\log_{b} (\frac{1 - 3 x}{x}) = 3 .

Raise b to the powers we see on each side. We get

b^{\log_{b} (\frac{1 - 3 x}{x})} = b^{3} .

The left-hand side simplifies greatly. We get

\frac{1 - 3 x}{x} = b^{3} .

The rest is elementary algebra. The above equation is (for

x \neq 0

) equivalent to

1 - 3 x = b^{3} x,

which is an easily solved linear equation.
Comment: There was no need to do the preliminary manipulation. We are told that

\log_{b} (1 - 3 x) = 3 + \log_{b} x .

Raise b to the power on the left-hand side, the right-hand side. We obtain

b^{\log_{b} (1 - 3 x)} = b^{3 + \log_{b} x} .

By the "laws of logarithms" this yields

1 - 3 x = b^{3} x .

Answers (1)