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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.

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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 7:43
In any event, remember that $\log_ab=c$ is just another way to write $a^c=b$. – Gerry Myerson Feb 27 '13 at 7:44
it is how to find the x. – cdummy Feb 27 '13 at 7:45
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. – Rahul Feb 27 '13 at 7:52
@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. – Arjang Feb 27 '13 at 12:18
up vote 2 down vote accepted

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.

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so the value of x is? – cdummy Feb 27 '13 at 7:46
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 7:52
@BabakS. Gerry was talking to cdummy. – zaarcis Feb 27 '13 at 9:04
I have never seen the argument of $\log$ written with a superscript. Shouldn't it be $\log_a b$ instead of $\log_a^b$? – Rahul Feb 27 '13 at 9:29
There is no need to apologize! Everyone has the right to post an answer here. – Rahul Feb 27 '13 at 9:47

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.

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Not much different from what @Babak.S wrote, yet efficient. – Parth Kohli Feb 27 '13 at 10:47

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