# 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?

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

• I checked the book and the first expression is $\sqrt{1+log_{0.04}x}$ – sku Mar 8 '18 at 16:17
• Yes, you are right, it should be 0.04. Does it become simpler as V.Asnin said? – James Warthington Mar 8 '18 at 18:46

## 3 Answers

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 $x$.

• I thank you very much for your instructive solution, also thank you for rechecking your answer. The equation only satisfies for t=1, which means x=1/0.04=25, which matches the answer in the back of the book. Thank you again! – James Warthington Mar 9 '18 at 1:09

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.

• I was in the middle of typing the same thing as a comment when this popped up. – saulspatz Mar 6 '18 at 4:50

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.

• @Leibovici: Thank you for your answer, I like your suggestion very much. But I have pursued this several ways and cannot reduce the expression into anything simpler. The result that I have come up with is simply $t=\frac{3+t\beta-2\sqrt{3+t\beta}}{\alpha }$ – James Warthington Mar 6 '18 at 23:20
• If I let $t=\sqrt{3+log_bx}$ where $b=0.04$ then $x=0.04^{t(t+2)}$. This is obviously a complicated expression. Do you think I can use derivative to examine the root of this equation? – James Warthington Mar 6 '18 at 23:26