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$$\log_{27}{8(\log_x{3})} = 1 $$ Please provide any quick method to solve this kind of problems.
The above is just an example.
Any better and tough examples with explanation could also be fine.

jimjim
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cdummy
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    Is it really nested? $\log_{27}(8\log_x3)=1$? Not $(\log_{27}8)(\log_x3)=1$? – Gerry Myerson Feb 27 '13 at 07:43
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    In any event, remember that $\log_ab=c$ is just another way to write $a^c=b$. – Gerry Myerson Feb 27 '13 at 07:44
  • it is nested.so how to find the x. – cdummy Feb 27 '13 at 07:45
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    Read what Gerry wrote and think for yourself. $\log_a b=c$ means that $a^c=b$. What can you put in for $a$, $b$, and $c$ to make it fit your problem? Forget the nesting for a moment and just concentrate on the outer logarithm. –  Feb 27 '13 at 07:52
  • @Ethereal : Thanks for the edit, but putting the latex before English is intentional, the Related Links on right hand side will display more. This is a case of visibility of relavent information to having some english words in the right order. – jimjim Feb 27 '13 at 12:18

2 Answers2

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If $\log_{a} b=c,~~b>0,a>0,a\neq 1$ then note that we have $$a^c=b$$ So assuming your equation; we have $$8\log_{x}{3}=27^1=27\Longrightarrow\log_x3=\frac{27}{8}=\Big(\frac{3}{2}\Big)^3$$ or $$8\log_{x}{3}=27\Longrightarrow\log_x{3^8}=27$$

Edit: I used $\log_{a}^b$ wrongly insted of $\log_a b$ and the following is due to this mistake. Apology and Excuse.

Mikasa
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  • so the value of x is? – cdummy Feb 27 '13 at 07:46
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    Can you not do anything with logarithms yourself? You've seen the formulas, you've seen them applied --- I'm sure you can learn enough from what's been done to continue the work. – Gerry Myerson Feb 27 '13 at 07:52
  • @GerryMyerson: Ok Gerry. sorry if I made mistakes here. I remove it soon. Forgive. – Mikasa Feb 27 '13 at 07:54
  • @Babak S. why you edited the previous answer. What is wrong with the previous answer – cdummy Feb 27 '13 at 07:57
  • @cdummy: Gerry is a Master here, and when he noted me some of my faults I wanted to obey him. If it is possible allow me to remove my answer. I am sure it makes my feelings good here. Maybe, I should learn how to treat here better. Thanks – Mikasa Feb 27 '13 at 08:01
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    @BabakS. Gerry was talking to cdummy. – zaa Feb 27 '13 at 09:04
  • @zaarcis: Yes I found out that, but it was late. You know the page is freeze for a while when you are writing any post especially when you are doing an answer. Indeed, the page needs a refresh till you find some other points are revealed like comments and anothe answers. I don't want to annoy anyones here. :-( – Mikasa Feb 27 '13 at 09:11
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    I have never seen the argument of $\log$ written with a superscript. Shouldn't it be $\log_a b$ instead of $\log_a^b$? –  Feb 27 '13 at 09:29
  • @ℝⁿ.: What can I do. I didn't note that you and Gerry were taking to the OP theaching him/her to get the way by himsel/herself. I appogize you both if I interrupted your way unwanted. Sorry. I feel bad dear friend.... – Mikasa Feb 27 '13 at 09:35
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    There is no need to apologize! Everyone has the right to post an answer here. –  Feb 27 '13 at 09:47
  • @BabakS. It's okay. :) – zaa Feb 27 '13 at 09:54
  • Babak, I'm sorry I didn't make it clear that my comment was directed at OP, not at you. – Gerry Myerson Feb 27 '13 at 12:11
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Another possible way is noticing that your equation can be written as $\log_{27}(8 \log _x 3) = \log_{27}27 \quad \iff \quad \log_x 3^8 = 27$. Hint: Logarithmic to exponential form.

P.K.
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