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

Solve in terms of b:
${\mathrm{log}}_{b}\left(1-3x\right)=3+{\mathrm{log}}_{b}x$
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Gianna Travis
Remember, any $\mathrm{log}$ to the base b, such as ${\mathrm{log}}_{b}\left(y\right)$, is in its heart an exponent. Let us start, then, from your expression
${\mathrm{log}}_{b}\left(\frac{1-3x}{x}\right)=3.$
Raise b to the powers we see on each side. We get
${b}^{{\mathrm{log}}_{b}\left(\frac{1-3x}{x}\right)}={b}^{3}.$
The left-hand side simplifies greatly. We get
$\frac{1-3x}{x}={b}^{3}.$
The rest is elementary algebra. The above equation is (for $x\ne 0$) equivalent to
$1-3x={b}^{3}x,$
which is an easily solved linear equation.
Comment: There was no need to do the preliminary manipulation. We are told that
${\mathrm{log}}_{b}\left(1-3x\right)=3+{\mathrm{log}}_{b}x.$
Raise b to the power on the left-hand side, the right-hand side. We obtain
${b}^{{\mathrm{log}}_{b}\left(1-3x\right)}={b}^{3+{\mathrm{log}}_{b}x}.$
By the "laws of logarithms" this yields
$1-3x={b}^{3}x.$