Solve the equation: $x^{1/(1+\log x)}=10^3$

Question

Solve the equation: $x^{1/(1+\log x)}=10^3$.

I thought to take the logarithm on both sides but I couldn't find a solution.

Answer 1

I assume by $\log(x)$ you mean $\log_{10}(x)$. Taking $\log_{10}$ on both sides, we obtain $$\dfrac1{1+\log_{10}(x)} \cdot \log_{10}(x) = \log_{10}\left(10^3\right) = 3$$ Rearranging this, gives us $$\log_{10}(x) = 3 + 3\log_{10}(x) \implies \log_{10}(x) = -\dfrac32 \implies x = 10^{-3/2} = \dfrac1{10\sqrt{10}}$$

Answer 2

$$x^{1/(1+\log x)}=10^3$$

$$\log(x^{1/(1+\log x)})=\log(10^3)$$

$${\log(x) \over {1+\log(x)}}=3\log(10)=3$$ $$\log(x)=3 (1+\log(x))$$ $$\log(x)=3 +3\log(x)$$ $$-2\log(x)=3$$ $$\log(x)=-\frac{3}{2}$$ $$x=10^{-{3 \over 2}}$$

Answer 3

let us make a change of variable $$\log_{10}x = a, x = 10^a.$$ with that we have $$x^{\frac1{1+a}} = 10^3\to \frac1{1+a}\log_{10}x= 3,\frac{a}{1+a} = 3$$

solving for $a,$ we find that $$a = -2/3 \to x = 10^{-2/3}=\frac1{\sqrt[3]{100}}$$