I want to find the value of this expression in as simple way as possible.
$$ \frac{1+ 2\log_3 2}{(1+ \log_3 2)^2}+ \log^2_6 2 $$
I simplified and I am stuck at $$ \frac{1+2\log_3 2+2\log_6 2+4\log_3 2×2\log_6 2}{1+4\log_3 2} $$
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Formatting tip use \log for proper spacing also use ^n for raising something to nth power i.e x^2 for $x^2$. – kingW3 Jun 30 '17 at 16:49
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I simply expanded the denominator and found nothing helpful,it doesn't simplifies. I think I'm not doing it the right way – Fghj Jun 30 '17 at 16:50
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See now I have edited. – Fghj Jun 30 '17 at 17:02
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Welcome to MSE, as you see you need to write your efforts before asking here, because may some user downvote your question. In this Wikipedia you have how translate your logarithms as logarithms to base $e$, that are $\log_e x=\ln x$. Is only grammar. The formula that I am saying is in section Change of base. Was the answer below. +1 and good luck. – Jun 30 '17 at 17:07
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@user243301 A greatest of the great thanks to you. I'll never forget your guidelines in future. – Fghj Jun 30 '17 at 17:11
1 Answers
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Knowing that $\log_ab = \frac{\log b}{\log a}$
Then the expression becomes:
$$\frac{1 + 2\frac{\log 2}{\log 3}}{(1+\frac{\log 2}{\log 3})^2} + (\frac{\log 2}{\log 6})^2$$
$$= \frac{1 + 2\frac{\log 2}{\log 3}}{1+2\frac{\log 2}{\log 3} + \frac{\log^2 2}{\log^2 3}} + (\frac{\log 2}{\log 3 + \log 2})^2$$
$$= \frac{\log^23 + 2\log 2\log 3}{\log^23+2\log2\log 3 + \log^2 2} + \frac{\log^2 2}{\log^23 + 2\log2\log3 + \log^22}$$
$$= \frac{\log^23 + 2\log 2\log 3 + \log^22}{\log^23+2\log2\log 3 + \log^2 2}$$
$$=1$$
