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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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    You knew what to do, I think. There was carelessness about the notation for $\log_3$, and carelessness can turn an easy right answer into$a$wrong one. In the line in the original that says $\ln((6x+2)/x^3)=2^3$ the $\ln$ was probably unintended, since you recovered from it on the next line. The $2^3$ is probably a slip of the tongue for $3^2$, natural since one sees $2^3$ a lot more often than $3^2$.2011-11-11

4 Answers 4

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It's logarithm base 3, so your second line should read:

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

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.