A rational number which is 50 times its own logarithm to the base 10 is?

Question

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 **

Answer 1

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 $100$

Answer 2

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 $x=10^2$.

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There is another solution to the equation, but it is not rational.

Answer 3

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 $x=100$.

Answer 4

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, $x=100$.