So I was looking through the homepage of Youtube to see if there were any math equations that I thought that I might be able to solve when I came across this video by SyberMath finding the value of $x$ in $$2^{x^2}\cdot5^x=10$$ which I thought that I might be able to do. Here is my attempt at solving the aforementioned equation:$$2^{x^2}\cdot5^x=10$$$$2^{x^2}5^{x^2}5^{x^2-x}=10$$$$\dfrac{2^{x^2}5^{x^2}5^{x^2}}{5^x}=10$$$$\dfrac{50^{x^2}}{5^x}=10$$$$50^{x^2}=5^x10$$$$x^2\ln50=x\ln5+\ln10$$$$x\ln50=\ln5+\dfrac{\ln10}x$$$$x\ln50-\ln5=\require{cancel}\cancel{\ln5-\ln5}+\dfrac{\ln10}x-\ln5$$$$x\ln10=\dfrac{\ln2}x$$$$x^2\ln10=\ln2$$$$x^2=\dfrac{\ln2}{\ln10}\qquad\text{The reason being if I had subtracted that}$$$$\text{it would have resulted in }(x^2-1)\ln10=-\ln5$$$$x=\sqrt{\dfrac{\ln2}{\ln10}}$$
My question
Is the solution that I have arrived at correct, or what could I do to attain the correct solution more easily/attain it more quickly?
Mistakes I might have made
- Logarithms
- Exponentiation (ties into logarithms)
- Basically everything
Best Answer
$2^{x^2}5^x=10$
By inspection can see that 1 is a solution. Are there others?
Take the log of both sides (the base doesn't technically matter, but base 10 seems to be what the problem was designed for.)
$\log(2^{x^2})+\log (5^x) = 1\\ (\log 2)x^2 + (\log 5) x - 1 = 0$
Now we have a quadratic.
Rather than using the quadratic formula, we can use the fact that we know that 1 is a root. By Vieta's formulas, we know that $r_1r_2 = \frac {c}{a}$ and $r_1+r_2 = -\frac ba.$ And, the other root must be $-\frac {1}{\log 2} = -(\frac {\log 5}{\log 2} + 1) = -\log_2 10$