$$\log_{3x}81 = 2$$ How would I go about solving this? This is what I tried:
$$\log_{3x}81 = 2$$
$$\frac{\log81}{\log 3 + \log x }= 2$$
Where do I go from here?
If I isolate $\log x$ on one side, how do I get rid of the log?
$\endgroup$ 53 Answers
$\begingroup$$\log_{3x}(81)=2$ is equivalent to
$$(3x)^2=9x^2=81$$
by the definition of the logarithm.
$$9x^2=81 \Leftrightarrow x^2=9$$
This gives solutions $x=3$ and $x=-3$, but only $x=3$ is a solution, since the base of a logarithm must be greater than zero.
$\endgroup$ 1 $\begingroup$For any two real numbers $b$ and $x$ where $b$ is positive and $b ≠ 1$,
$$ y=b^z\Leftrightarrow z=\log_b(y) $$ so for $\log_{3x}81=2$ we have $$ (3x)^2=81=(3\cdot 3)^2\Rightarrow x=3 $$
$\endgroup$ $\begingroup$Aside from the simpler way of solving this that someone else has already posted, here's something else you can do: $$ \frac{\log81}{\log 3 + \log x }= 2 $$ $$ \frac{\log 3 + \log x}{\log 81} = \frac 1 2 $$ $$ \log 3 + \log x = \frac 1 2 \log 81 $$ $$ \log x = \frac 1 2 \log 81 - \log 3 = \frac 1 2 \log(3^4) - \log 3 $$ $$ =\frac 1 2\cdot 4 \log 3 - \log 3 = \left(\frac 1 2\cdot 4 - 1\right)\log 3 = \log 3. $$ Now you have $$ \log x = \log 3, $$ and since logarithmic functions are one-to-one, you can conclude that $x=3$.
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