I have two expressions: $\log_2 5$ and $\log_3 13$.
And I need to determine which expression is greater using exactly derivatives and not having a calculator.
I know that $(y = \log_2 5)'$ = $\frac{1}{5ln2}$ and $(y = \log_3 13)'$ = $\frac{1}{13ln3}$
But what are the further steps(algorithm) in such cases to determine what is greater?
Addendum added to respond to the comment of vortexsf24.
A very similar problem was posted on MathSE recently. Another reviewer found the following solution that I would not have thought of. If I remember correctly, it was Dietrich Burde; I could be mistaken about who found the solution. I can't find the original posting; it may have been deleted.
You can show, with minimal calculation (without a calculator) that
$$3 \times \log_2(5) < 7 < 3 \times \log_3(13).$$
This simplifies to showing that
$$\log_2\left(5^3\right) < 7 < \log_3\left(13^3\right).\tag1 $$
In (1) above, the LHS inequality is fairly easy. Simply notice that $~5^3 = 125 < 128 = 2^7.~$ So, since $~2^7 > 125,~$ you must have that $~\displaystyle \log_2\left(5^3\right)~$ must be $~< 7.$
In (1) above, the RHS inequality is more difficult, but can still be confirmed by pencil and paper (again without a calculator). Simply verify that
$$3^7 = 2187 < 2197 = 13^3.$$
So, since $~3^7 < 13^3,~$ you must have that $~\displaystyle \log_3\left(13^3\right)~$ is $~> 7.$
$\underline{\text{Addendum}}$
Responding to the comment of vortexsf24.
"I need any possible solution to the problems of such kind(if possible derivatives) excluding the way of making the bases equal and calculator"
My answer did not make the bases equal.
Also, it isn't really appropriate to try to use derivatives to attack this type of problem. Both sides of the inequality are constants. The derivative of a constant is $~0.~$
One approach for trying to force-fit the use of derivatives onto this problem is as follows:
The problem can be restated as :
Prove that
$$\frac{\log(5)}{\log(2)} < \frac{\log(13)}{\log(3)},$$
where $~\log(x)~$ refers to the natural logarithm.
This can be re-stated as
Prove that $$\frac{\log(5) \times \log(3)}{\log(2)} < \log(13). \tag2 $$
Then, using (2) above as the starting point of your analysis, you would set
$$f(x) = \log(x) - \frac{\log(5) \times \log(3)}{\log(2)}. \tag3 $$
It is easy to see that on $~\Bbb{R^+},~$ since $~f'(x) = \dfrac{1}{x} > 0,~$ the function $~f(x)~$ is strictly increasing.
Further, since $~\log(1) = 0,~$ you know that $~f(1) < 0.~$ Therefore, since $~\lim_{x\to\infty} \log(x) = \infty,~$ you also know that $~f(x)~$ must cross the $~x$-axis exactly once, for $~x \in (1, +\infty).$
So, the problem is reduced to determining whether the (sole) root of the equation $~f(x) = 0~$ is less than $~13,~$ equal to $~13,~$ or greater than $~13.~$
Personally, I don't see how this approach gets anywhere. You know that :
$\displaystyle f(1) = - \frac{\log(5) \times \log(3)}{\log(2)}.$
$f'(x) = \dfrac{1}{x}, ~f''(x) = \dfrac{-1}{x^2}, f'''(x) = \dfrac{2}{x^3}, \cdots $
I don't see any way of using these insights to determine whether the rate of increase of $~f(x)~$ is slowed down enough so that $~f(x)~$ crosses the $~x$-axis before $~x=13,~$ rather than after.