Why does this work? What is the best exponent to approximate the logarithm? Suppose i want to get a upper bound of $log_53$ without other tools than elementary algebra.
I can do this: 
Let $x = log_53$, then $5^x=3$
I raise both sides to $4$, then: $(5^x)^4 = 3^4$ And since $3^4 < 5^3$ it follows that $5^{4x} < 5^3$, then $x < \frac{3}{4}$ 
However, if i raise both sides to $13$ is equal to: $(5^x)^{13} = 3^{13}$ And since $3^{13} < 5^9$ it follows that $5^{13x} < 5^9$, then $x < \frac{9}{13}$
In this case with exponent $13$ the approximation is better.
So, with different exponent i can get different upper bounds. Similarly, the same can be done with the lower bound.
And my question is, 
Given $log_ab = x$, what is the best exponent $m$ such that $(a^x)^m = b^m$ gives the better approximation for both cases?(Lower and upper bounds) 

There's no "best exponent", because you can always get better and better exponents, but you can use continued fractions to do this systematically. With continued fractions, we will write $x$ in the form
$$
   x = a_0 + \cfrac1{a_1 + \cfrac1{a_2 + \cfrac1{a_3 + \cfrac1{a_4 + \dots}}}}.
$$
where $a_0, a_1, a_2, a_3, a_4, \dots$ is an infinite sequence of natural numbers. 
We can truncate this sequence at any point, getting upper or lower bounds depending on where we stop. (If you stop at an even $a_k$, you get a lower bound; if you stop at an odd $a_k$, you get an upper bound.)
We compute the sequence iteratively: $a_0$ is $\lfloor x\rfloor$ (the greatest integer less than $x$) and to find $a_1$ onward you just replace $x$ by $\frac{1}{x - a_0}$. 

All this is abstract and general, so here is how it will go in your example.
Let $x = \log_5(3)$. Then we can estimate that $0 < x < 1$ (because $5^0 < 3 < 5^1$) so $a_0 = \lfloor \log_5(3)\rfloor = 0$, and we will keep going with $\frac1{\log_5(3)} = \log_3(5)$.
We can estimate that $1 < \log_3(5) < 2$, because $3^1 < 5 < 3^2$, so $a_1 = \lfloor \log_3(5)\rfloor = 1$, and we will keep going with $\frac{1}{\log_3(5)-1} = \frac1{\log_3(5/3)} = \log_{5/3}(3)$.
We can estimate that $2 < \log_{5/3}(3) < 3$, because $(\frac53)^2 < 3 < (\frac53)^3$ (this expands to $25 < 27$ but $81 < 125$), so $a_2 = \lfloor \log_{5/3}(3) \rfloor = 2$, and we will keep going with $\frac1{\log_{5/3}(3) - 2} = \frac1{\log_{5/3}(27/25)} = \log_{27/25}(\frac53)$.
We could estimate that $6 < \log_{27/25}(\frac53) < 7$, because $(\frac{27}{25})^6 < \frac53 < (\frac{27}{25})^7$, but this has gotten tricky to check: it involves checking that $3^{19} < 5^{13}$ but $3^{22} > 5^{15}$. This gives us $a_3 = 6$, but maybe we'd better quit while we're ahead, or get a computer.
At this point, our best lower bound is
$$
    a_0 + \cfrac1{a_1 + \cfrac1{a_2}} = 0 + \cfrac1{1 + \cfrac12} = \frac23
$$
and our best upper bound is
$$
   a_0 + \cfrac1{a_1 + \cfrac1{a_2 + \cfrac1{a_3}}} = 0 + \cfrac1{1 + \cfrac1{2 + \cfrac16}} = \frac{13}{19}.
$$
