<s>[INST] How do you solve this equation $\sqrt{1+log_{0.004}x}+\sqrt{3+log_{0.2}x}=1$? 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?
$\sqrt{1+log_{0.004}x}+\sqrt{3+log_{0.2}x}=1$
 [/INST] 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 $x$. 
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