A rational number which is 50 times its own logarithm to the base 10 is? This question is from Advanced problems in mathematics for jee . I got it as a challenging question.
I tried it in this way
    50 log x base 10 = x
But there seemed no solution for it as per my level.
**Please don't do it based on option. I want solution as if there was no option given **
 A: We have that $50\log_{10}(x)=x$. Therefore $10^{x}=x^{50}$. Let $x=p/q$, with $(p,q)=1$. Then $$q^{50q}10^{p}=p^{50q}$$
It follows that $q=1$ and $p$ is a power of $10$ (and therefore so is $x$). 
Put $x=10^n$. Then $50n=10^n$. Since $2$ and $5$ divide $10$ to the same power, we must have $2|n$. Put $n=2k$. Then $$100k=100^k$$
We see that $k=1$ satisfies the equation, but for $k>1$ we have that $100^k\gg100k$.
Therefore $x=10^2$.

There is another solution to the equation, but it is not rational.
A: The solution is 100. Just write the equation:
$$x=50\log x$$
For $x=1$ you have $1=50\times 0$ which is false.
For $x=10$ you have $10=50\times 1$ which is false.
For $x=100$ you have $100=50\times 2$ which is true.
For $x=1000$ you have $1000=50\times 3$ which is false.
If you don't want to check the options the general way to solve this equation is:
$x=50\frac {\ln x} {\ln 10}$
$\frac x {\ln x}=\frac {50} {\ln 10}$
$\frac {\ln x} x=\frac {\ln 10} {50}$
$e^{-\ln x}\ln x=\frac {\ln 10} {50}$
$-e^{-\ln x}\ln x=-\frac {\ln 10} {50}$
Now you apply Lambert's W function which is defined as $W(z)e^{W(z)}=z$
$-\ln x=W(-\frac {\ln 10} {50})$
$$x=e^{-W(-\frac {\ln 10} {50})}$$
Which is equal to $100$
A: We have
$$x = 50\log_{10}x \implies x = \log_{10}\left(x^{50}\right) \implies 10^x = x^{50} \implies 2^x \cdot 5^x = x^{50}$$
Hoping that $x$ is an integer, we then see that $x$ has only $2$ and $5$ as its prime divisors, i.e., $x = 2^a 5^b$. This means we have
$$2^x \cdot 5^x = 2^{50a} \cdot 5^{50b}$$
This means $x=50a=50b \implies a=b$, i.e., $x=2^a \cdot 5^b = 10^a$. Further, $50 \mid x$, which means $a \geq 2$. Hence, plugging in $x=10^a$ in $x=50\log_{10}x$, we obtain
$$10^a = 50a$$
Note that for $a>2$, we have $10^a-50a > 0$. Hence, $a=2$ is the only solution, which gives us $x=100$.
A: Here's one way of thinking about it.  The equation is $x=50\log_{10}x$.  Suppose
$x=10^n$.  We need to make some kind of guess to get started.  Might has well gues something we at least will have a chance of solving.
Substituting $10^n$ for $x$ gives
$$10^n=50\log_{10}10^n\implies 10^n=50n$$
So, we need a multiple of $50$ which is also a power of $10$.  The only $n$ which will work are of the form $2\cdot 10^k$.  So,
$$10^{2\cdot 10^k}=50\cdot2\cdot10^k=10^{k+2}$$
We are now reduced to solving,
$$2\cdot10^k=k+2$$
$k=0$ clearly works, but for larger values of $k$ $2\cdot10^k$ will be larger than $k+2$.
If $k=0$, $x=2\cdot 10^0=2$.  Therefore, $x=100$.
