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

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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    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}.$2011-12-06

1 Answers 1

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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 $1-3x=b^3 x.$