Yet another logarithm multiplication doubt

I'm here again, now with a doubt on multiplication on logarithms, I have the expression: $(\log_5 2 + \log_5 3 \cdot\log_3 4) \cdot \log_2 5$

I've evaluated it to: $$\left(\log_5 6 \cdot \dfrac{\log_5 4}{\log_5 3}\right) \cdot \dfrac{1}{\log_5 2}$$

What should I do with this multiplications ?

EDIT: So, I failed miserably on algebra through, so I evaluated now to:

$= (\log_5 2 + \log5 4) * \log_2 5$

$= (\log_5 8)*\dfrac{1}{\log+5 2}$

$= \log_5(\frac{8}{2}$)

$= \log_5 4$

So, do I solved it correctly now ? Thank you. EDIT 2:

$= (\log_5 2 + \log_5 4) * \log_2 5$

$= (\log_5 8)*\dfrac{1}{\log_5 2}$

$= \dfrac{\log_5 8}{\log_5 2}$

$= \log_2 8 = 3$

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You can't get to that expression. Notice that $a+b\cdot c\neq(a+b)\cdot c$. What you should do is: \begin{align*}(\log_5 2 + \log_5 3 \cdot \log_3 4) \cdot \log_2 5&=\left(\log_5 2 + \log_5 3 \cdot \frac{\log_5 4}{\log_53}\right) \cdot \frac{1}{\log_52}\\ &=\left(\log_5 2 + \log_5 4\right) \cdot \frac{1}{\log_52}=\frac{\log_58}{\log_52}\\ &=\log_28=3\end{align*} You had another mistake: $\frac{\log_ab}{\log_ac}\neq\log_a\frac bc$

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Thanks, I've corrected this mistake in the edit –  aajjbb Nov 19 '12 at 0:09
I've interpreted the property wrongly, but, I didn't knew how did you passed $\dfrac{\log_5 8}{\log_5 2}$ to $\log_2 8 = 3$ –  aajjbb Nov 19 '12 at 13:41
It is the same property you used in the first step: $\frac{\log_ab}{\log_ac}=\log_cb$ –  Dennis Gulko Nov 19 '12 at 14:38

You have an algebra error: $\log_52+\log_53\cdot\log_34\ne\log_56\cdot\log_34$. It should be

\begin{align*} (\log_52+\log_53\cdot\log_34)\log_25&=\left(\log_52+\log_53\cdot\frac{\log_54}{\log_53}\right)\frac1{\log_52}\\ &=1+\frac{\log_54}{\log_52}\\ &=1+\frac{\log_52^2}{\log_52}\;; \end{align*}

can you finish it from there?

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