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

$$\log_b (1 - 3x) = 3 + \log_b x$$

If I use the properties of logs, I end up with

$$\log_b \left(\frac{1 - 3x}{x}\right) = 3$$

From there, the example I have says to exponentiate both sides, however, they use a $\log_2(\text{equation}) = 2$. They then raise both sides to the $2$. I can't seem to do this here, does someone have an example that better fits my situation?

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Raise $b$ to the powers involved. Note that $b^{\log_b(y)}=y$. So you should end up with $\frac{1-3x}{x}=b^3$. – André Nicolas Dec 6 '11 at 20:16
I've got it solved, thanks. If you answer this question with your comment, I'll mark this as the accepted answer. – erimar77 Dec 6 '11 at 20:31
It seems what is really missing here is an understanding of logarithms: Recall that $$\log_b(z) = y \quad\iff\quad b^y = z.$$ In other words, $\log_b(z)$ is the power to which you exponentiate $b$ to get $z$. In your case, $$\log_b\left( \frac{1-3x}{x} \right) = 3 \quad\iff\quad b^3 = \frac{1-3x}{x}.$$ – JavaMan Dec 6 '11 at 20:58

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 $$\log_b\left(\frac{1-3x}{x}\right)=3.$$ Raise $b$ to the powers we see on each side. We get $$b^{\log_b\left(\frac{1-3x}{x}\right)}=b^3.$$ 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 $$\log_b(1-3x)=3+\log_b x.$$ Raise $b$ to the power on the left-hand side, the right-hand side. We obtain $$b^{\log_b(1-3x)}=b^{3+\log_b x}.$$ By the "laws of logarithms" this yields $$1-3x=b^3 x.$$