Find the value of $x$ that satisfies the equation $\log_{10} \left(\frac{x^{\frac{1}{x}}}{x^{\frac{1}{x+1}}}\right) = 1/5050$ .

Question

I tried it many times and it went bit of lengthy , i reached until

\begin{equation*} \log_{10}(x^{1/(x^2+x)}) \end{equation*}

then i multiplied $2$ both numerator and denominator and then it is consecutive terms in denominator I got my answer $100$. Just want to know if it is correct and a shorter way to solve it ?

Answer 1

This is equivalent to $$\log(x^{\frac{1}{x(1+x)}})=\frac{1}{5050}$$ Then:

$$x^{\frac{1}{x(1+x)}}=10^{\frac{2}{100*101}}$$ Or $$x^{\frac{1}{x(1+x)}}=100^{\frac{1}{100*101}}$$

Thus $x=100$.

Answer 2

Given

$$\log_{10}\left(\frac{x^\frac 1x}{x^{\frac 1{x+1}}}\right)={1\over 5050}$$

the first couple important transformations are to note $\frac{x^\frac 1x}{x^{\frac 1{x+1}}}=x^{\frac 1{x^2+x}}$, and then that we can rearrange like so:

$$\frac 1{x^2+x}\log_{10} x=\frac 1{5050}\\ \to \log_{10}(x^{5050})=x^2+x\\ \to x^{5050}=10^{x^2+x}$$

Assuming that $x$ is positive, it makes sense to guess that $x=10^k$, yielding

$$10^{5050k}=10^{10^{2k}+10^k}\\ 5050k=10^{2k}+10^k\\ 10100m=10^{4m}+10^{2m}=10000^m+100^m$$

We can go from "assuming that $x$ is positive" to knowing that it must be (for real $x$), since negative $x$ in $x^{1/{(x^2+x)}}$ is complex ($x^2+x$ is strictly even); therefore, our substitution $x=10^k=100^m$ is legitimate. Trying the first couple integers $k$ yields a solution $m=1,k=2, x=100$. A second real-valued solution exists, by noting that $5050k$ changes from greater than $10^{2k}+10^k$ to less than this quantity as $k$ goes from $1$ to $0$.