# Logarithmic Equation: Solve for $x$

$$\log_{3x}81 = 2$$ 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?

• $81=3^4$, so $\log81=4\log3$. – Akiva Weinberger Nov 18 '14 at 17:57
• Or, do $(3x)^\textrm{both sides}$ from the start. – Akiva Weinberger Nov 18 '14 at 17:59
• Do you mind showing me your steps for this? Are you starting at where I left off? – McB Nov 18 '14 at 18:00
• The second thing is starting from the start. Remember that $(3x)^{\log_{3x}(\textrm{stuff})}=\textrm{stuff}$. – Akiva Weinberger Nov 18 '14 at 18:01
• Aside from the simple way of doing this that others have already posted, you can also solve the equation $\dfrac{\log81}{\log 3 + \log x }= 2$ for $x$. I explained how in an answer posted below. ${}\qquad{}$ – Michael Hardy Nov 18 '14 at 18:16

$\log_{3x}(81)=2$ 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.

• Very helpful, thanks. For some reason I failed to see that I could isolate x very easily once I "booted the log". – McB Nov 18 '14 at 18:06

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 $$(3x)^2=81=(3\cdot 3)^2\Rightarrow x=3$$

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 $x=3$.

• Thank-you, that is very helpful. – McB Nov 18 '14 at 18:18