Nice exponential equation by SyberMath: $2^{x^2}\cdot5^x=10$

Question

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:$$2^{x^2}\cdot5^x=10$$$$2^{x^2}5^{x^2}5^{x^2-x}=10$$$$\dfrac{2^{x^2}5^{x^2}5^{x^2}}{5^x}=10$$$$\dfrac{50^{x^2}}{5^x}=10$$$$50^{x^2}=5^x10$$$$x^2\ln50=x\ln5+\ln10$$$$x\ln50=\ln5+\dfrac{\ln10}x$$$$x\ln50-\ln5=\require{cancel}\cancel{\ln5-\ln5}+\dfrac{\ln10}x-\ln5$$$$x\ln10=\dfrac{\ln2}x$$$$x^2\ln10=\ln2$$$$x^2=\dfrac{\ln2}{\ln10}\qquad\text{The reason being if I had subtracted that}$$$$\text{it would have resulted in }(x^2-1)\ln10=-\ln5$$$$x=\sqrt{\dfrac{\ln2}{\ln10}}$$

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My question

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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?

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Mistakes I might have made

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1. Logarithms
2. Exponentiation (ties into logarithms)
3. Basically everything

Answer 1

$2^{x^2}5^x=10$
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 $-\frac {1}{\log 2} = -(\frac {\log 5}{\log 2} + 1) = -\log_2 10$

Answer 2

Quite a neat way is to write $10=2\cdot5$ and divide both sides by this, getting $$2^{x^2-1}\cdot5^{x-1}=1,$$ or $$\left(2^{x+1}\cdot 5 \right)^{x-1}=1.$$ Then either $x=1$ or $2^{x+1}\cdot 5=1$. The latter is $2^x=\frac{1}{10}$, or $x=-\dfrac{\ln10}{\ln2}$.

Answer 3

\begin{align} 2^{\log_2(10)}=10=2^{x^2}5^x=2^{x^2}2^{\log_2(5)x}=2^{x^2+\log_2(5)x} &\Longrightarrow \log_2(10)=x^2+\log_2(5)x\\ &\Longrightarrow x^2+\log_2(5)x-\log_2(10)=0. \end{align}

Now, the solution is by completing squares.

Answer 4

The quadratic equation shown by other posters also comes about from applying the properties of exponents: $$ 2^{x^2}·5^x \ \ = \ \ 2^{x·x}·5^x \ \ = \ \ (2^x)^x·5^x \ \ = \ \ (2^x·5)^x \ \ = \ \ 10 $$ $$ \Rightarrow \ \ x·\log_a(2^x·5) \ \ = \ \ x·(\log_a 2^x \ + \ \log_a 5) \ \ = \ \ x·(x·\log_a 2 \ + \ \log_a 5) $$ $$ = \ \ (\log_a 2)·x^2 \ + \ (\log_a 5)·x \ \ = \ \ \log_a 10 \ \ . $$

This can then be factored as $$ (\log_a 2)·x·x \ + \ \left(\log_a \frac{10}{2} \right)·x \ − \ \log_a 10 $$ $$ = \ \ x·(\log_a 2)·x \ + \ (\log_a 10)·x \ − \ (\log_a 2)·x \ − \ \log_a 10 $$ $$ = \ \ (x \ − \ 1)·( \ [\log_a 2]·x \ + \ \log_a 10 \ ) \ \ = \ \ 0 \ \ . $$

We should expect the original equation to have two roots. The function $ \ a^{x^2} \ \ , \ a \ > \ 1 \ \ $ is an even function with a global minimum at $ \ (0 \ , \ 1) \ \ ; \ $ so, for instance, $ \ 2^{x^2} \ = \ 10 \ \Rightarrow \ x \ = \ \pm \sqrt{\frac{\log_a 10}{\log_a 2}} \ \ . \ $ The function used in our equation can be written as $$ 2^{x^2}·(2^{\log_2 5})^x \ \ = \ \ 2^{x^2 \ + \ \log_2 5·x} \ \ = \ \ 2^{( \ x \ + \ [\log_2 5]/2 \ )^2}·2^{-[\log_2 5]^2/4} \ \ , $$ which "shifts" $ \ 2^{x^2} \ $ "to the left" by $ \ \frac{\log_2 5}{2} \ \approx \ 1.1610 \ $ and "compresses it vertically" by a factor of $ \ 2^{-[\log_2 5]^2/4} \ \approx \ 2^{-1.3478} \ \approx \ 0.3929 \ \ . \ $ The function curve retains the convex ("concave upward") shape of $ \ 2^{x^2} \ \ $ with the indicated axis of symmetry and global minimum; it also continues to have a $ \ y-$intercept $ \ (0 \ , \ 1) \ \ . \ $ The equation $ \ 2^{x^2}·5^x \ = \ c \ $ thus has two solutions, one positive and one negative, for $ \ c \ > \ 2^{-[\log_2 5]^2/4} \ \ . $

The roots for $ \ c \ = \ 10 \ $ would be found from this expression as $$ \left( \ x \ + \ \frac{\log_2 5}{2} \ \right)^2 \ - \ \frac{(\log_2 5)^2}{4} \ \ = \ \ \log_2 10 $$ $$ \Rightarrow \ \ \left( \ x \ + \ \frac{\log_2 5}{2} \ \right)^2 \ \ = \ \ \frac{4·\log_2 10 \ + \ (\log_2 5)^2}{4} \ \ = \ \ \frac{4·[ \ \log_2 5 \ + \ \log_2 2 \ ] \ + \ (\log_2 5)^2}{4} $$ $$ = \ \ \frac{(\log_2 5)^2 \ + \ 4· \log_2 5 \ + \ 4 \ }{4} \ \ = \ \ \frac{ ( \ \log_2 5 \ + \ 2 \ )^2 }{4} $$

$$ \Rightarrow \ \ \ x \ + \ \frac{\log_2 5}{2} \ \ = \ \ \pm \left( \frac{ \log_2 5 \ + \ 2 }{2} \right) \ \ \Rightarrow \ \ x \ \ = \ \ 1 \ \ , \ \ -1 \ - \ \log_2 5 \ \ \approx \ \ -3.3219 \ \ . $$

Answer 5

Take $\ln$ of both sides and obtain the quadratic equation $$(\ln2)x^2+(\ln5)x-\ln 10=0.$$ Then solve by factoring $$(x-1)(x\ln2+\ln10)=0.$$