# Stuck with logarithm; Find the $x$ value

I am trying to find the value of $x$ ... but I'm absolutely stuck, some hints would be appreciated!

$$\log_3 (6x+2) - 2\log_3 (x)=2$$

My work so far:

\begin{align*} \log\left(\dfrac{6x+2}{x^2}\right) &= 2^3\\ 6x+2 &= 8x^2\\ -8x^2+6x+2 &= 0\\ x = 1,&\quad x = \dfrac{1}{4} \end{align*}

But I've tested with 1 and 1/4 and it's never equals to 2...

Where's my error ?

Thanks !

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Note: $\ln$ is usually reserved for the natural logarithm. To talk about logarithms of base $a$, we use $\log_a$, not $\ln_a$. – Arturo Magidin Nov 11 '11 at 20:29
Note that $\ln$ is reserved for the natural logarithm. If you want to specify a base, write $\log_3(\cdots)$. – Henning Makholm Nov 11 '11 at 20:29
Thanks, I've modified it – Jonathan Nov 11 '11 at 20:31
@HenningMakholm: Snap! – Arturo Magidin Nov 11 '11 at 20:31
@Jonathan: That second line is still wrong; that should be either $\log_3(...) = 2$, or drop the logs and have $3^2$ (not $2^3$). – Arturo Magidin Nov 11 '11 at 20:32

$$\dfrac{6x+2}{x^2} = 3^2$$

Note also that you have exponentiated, so there should be no $\ln$ anymore.

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You can't just drop the $\log_3$, so it should be $\log_3 \left( \frac{6x+2}{x^2} \right) =2$ which leads us to $6x+2=x^2 \cdot 3^2 =9x^2$.

Now you can go on the way you did: $-9 x^2+6x+2=0$ which is the case if and only if $x= \frac{1}{3} \pm \frac{\sqrt{3}}{3}$

Edit: I replaced $\ln$ by $\log_3$ in my answer and subsequently changed the results of the calculation.

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It is when $$\mathrm{log}_3 (6x+2) - 2 \mathrm{log} (x) = 2 \longrightarrow \log_3 \left( \frac{ 6x+2}{x^2} \right) = 2 \longrightarrow \frac{6x+2}{x^2} = 3^2 = 9.$$ When you get rid of logarithms you must take "3 to the both sides".

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I don't know how you got the first line; you should have $$\log_3\left(\frac{6x+2}{x^2}\right) = 2,$$ not the natural log and not equal to $2^3$.

Then you would conclude from this that $$3^{2} = \frac{6x+2}{x^2}.$$ Note in particular that you get $3^2$, not $2^3$. The logarithm base $3$ is what the exponent needs to be, not what the base needs to be.

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