How do you solve this equation $\sqrt{1+log_{0.004}x}+\sqrt{3+log_{0.2}x}=1$?

Question

This is a problem from Prilepko's book on Mathematics for High School, I have tried to solve this equation but cannot succeed. The hint is to put $t =$ $\sqrt{3+log_{0.2}x}$. But then it reduces to some quite complicated expression. If I've also tried $t=$ $\sqrt{1+log_{0.004}x}$ but that doesn't simplify anything. Could you help me on this?

$\sqrt{1+log_{0.004}x}+\sqrt{3+log_{0.2}x}=1$

Answer 1

Let $t = \sqrt{3+log_{0.2}x} \implies \frac{\ln x}{\ln 0.2} = t^2 - 3$

Now $\sqrt{1+log_{0.04}x} = \sqrt{1 + (t^2 - 3)\ln{0.2}/\ln{0.04}} = \sqrt{1 + \frac{t^2 - 3}{2}} = \sqrt{\frac{t^2 - 1}{2}} $

So we get $\sqrt{\frac{t^2 - 1}{2}} + t = 1 \implies t^2 -1 = 2(t-1)^2$

$ t^2 -1 = 2t^2 - 4t + 2 \implies t^2 - 4t + 3 = 0 \implies t = 3, 1$

solve for $x$.

Answer 2

It looks like a misprint. Were it $\sqrt{1+log_{0.04}x}$ instead of $\sqrt{1+log_{0.004}x}$ it would be much easier.

Answer 3

Make the problem more general $$\sqrt{1+\log_{a}x}+\sqrt{3+\log_{b}x}=1$$ and go to natural logarithms to get $$\sqrt{1+\frac{\log (x)}{\log (a)}}+\sqrt{3+\frac{\log (x)}{\log (b)}}=1$$ Now, for more simplicity, let $$t=\log(x)\qquad \alpha=\frac{1}{\log (a)}\qquad \beta=\frac{1}{\log (b)}$$ to end with $$\sqrt{1+\alpha t}+\sqrt{3+\beta t}=1$$ and use squaring.