# What is the product of the solutions to the equation (√10)(x^(log (x))) = x^2? [closed]

I have spent a few hours on this question but I can't seem to grasp it. I have found that taking the log of both sides doesn't give me any progress. The log exponent identity didn't work. I also tried moving both terms to one side and possible using vieta's but nothing worked. Lastly, I tried squaring both sides, but I don't think that brings me anywhere. Does the radical turning into 10^(1/2) have to do with anything. Can I have a few hints and could someone help me solve this equation. Thanks. I have found this question from a qualifying test that my middle school has for ARML.

• Just out of curiosity, when you say "taking log of both sides", exactly what did you get as a result of that before you gave up? What obstructed the log exponent identity from working? Commented Jul 12 at 20:12
• Can you please format your expressions using MathJax. It's hard to read otherwise. Commented Jul 12 at 20:15
• When I took the log of both sides of (√10)(log(x))(log(x))=2*log(x), I divided both sides by log(x) and kept account for division by 0, but I'm pretty sure that doesn't remove any solutions to get the equation (√10)(log(x)=2. Then proceeded to get (log(x)=(√10)/5, but after finding the solutions, the product doesn't match up with the answer. Commented Jul 12 at 20:22
• Yes it is base 10. When I did that and matched exponents, I got t^t+1/2=2t. But, I am unsure how to proceed. Commented Jul 12 at 20:35
• No. $\left( 10^t \right)^t = 10^{t^2}$ In general $\left( a^b \right)^c = a^{bc}$ Commented Jul 12 at 21:04

COMMENT.-$$\sqrt{10}x^{\log(x)}=x^2\iff x^{\frac{\log(\sqrt{10})}{\log(x)}}\cdot x^{\log(x)}=x^2$$ equating exponents we get $$\frac{\log(\sqrt{10})}{\log(x)}+\log(x)=2$$ so the quadratic equation $$X^2-2X+\log(10)=0$$ where $$X=\log(x)$$. The two solutions are $$\log(x)=\dfrac{-2\pm\sqrt2}{2}\Rightarrow x=10^{(\frac{-2\pm\sqrt2}{2})}$$
• The equation is $X^2-2X+\log(\sqrt{10})=0$, not $X^2-2X+\log(10)=0$. And the solutions are $X=\frac{\color{red}+2\pm\sqrt2}2$. And why does this answer begin with "Comment"? Commented Jul 12 at 22:02