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I tried using logarithms to find the answer to $x^2+5^x-10=0$ but I didn't have any luck. Is there a way of solving the above equation algebraically, or do you have to use a graphical method?

Joe
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    Things of the form $A^x + x^2 + C =0$ involve the Lambert $W$ (read: graphical). 'Traditional' analytical methods are insufficient. – AlvinL Feb 01 '20 at 12:29
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    @Joe: You can use analytic methods to show there are exactly two real roots. Then, you can use a graph to identify them and use numerical methods like Newton's Method to find them. – Moo Feb 01 '20 at 12:32
  • It is possible to derive that the negative root of this expression can be approximated by the root of the polynomial $$x^{3}-3x^{2}+3x+61=0$$, where the real root is actually; $x=1-\sqrt[^3]{62}\simeq-2.9579..$. Not a very good approximation, but something that can be done without much numerical calculations. – Mourad Feb 01 '20 at 15:33
  • @Moo How can I show that there are two real roots? – Joe Feb 01 '20 at 15:34
  • you can also see if you take $x=-\sqrt{10}$ Then $$(-\sqrt{10})^2+5^{-\sqrt{10}}-10=5^{-\sqrt{10}} \simeq 0.00616$$. – Mourad Feb 01 '20 at 15:48
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    @Joe: https://math.stackexchange.com/questions/2594196/how-to-show-that-an-equation-has-exactly-two-solutions – Moo Feb 01 '20 at 16:19
  • @Moo Thanks for sending me the link. I just researched the Lambert $W$. Does this count as a graphical function? Is there any way of algebraically finding the inverse of $f(x)=xe^x$? – Joe Feb 01 '20 at 16:38

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You are looking for the zero's of function$$f(x)=x^2+5^x-10$$ which is highly transcendental. The solutions would require numerical methods or special functions. Consider its derivatives $$f'(x)=2 x+5^x \log (5)\qquad \text{and} \qquad f''(x)=5^x \log ^2(5)+2 \quad >0 \quad \forall x$$ The solution of $f'(x)=0$ is given in terms of Lambert function $$x_*=-\frac{1}{\log (5)}W\left(\frac{1}{2}\log ^2(5)\right)\approx -0.413583$$ For this value $f(x_*)\approx -9.31500$ and this is a minimum (by the second derivative test); then two roots on each side of $x_*$.

Let me be lazy, and consider the expansion of $f(x)$ around $x=0$. This will give

$$f(x)=-9+x \log (5)+x^2 \left(1+\frac{\log ^2(5)}{2}\right)+O\left(x^3\right)$$ giving $$x_{1,2}=-\frac{\log(5)\pm\sqrt{36+19 \log ^2(5)}}{2+\log ^2(5)}$$ probably not very accurate but sufficient to start Newton method. The iterates will be $$\left( \begin{array}{cc} n & x_n \\ 0 & -2.3616508 \\ 1 & -3.3004059 \\ 2 & -3.1642570 \\ 3 & -3.1613032 \\ 4 & -3.1613018 \end{array} \right)$$

$$\left( \begin{array}{cc} n & x_n \\ 0 & 1.6604150 \\ 1 & 1.3887413 \\ 2 & 1.3171449 \\ 3 & 1.3131011 \\ 4 & 1.3130890 \end{array} \right)$$

Claude Leibovici
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$x^2+5^x-10=0$

$x= ± \sqrt{10-5^x}$

$10-5^x=1, 4, 9$

$10-5^x=1$ ⇒ $x=\frac{log (9)}{log (5)}=1.36$

$10-5^x=4$ ⇒ $x=\frac{log (6)}{log (5)}=1.11$

$10-5^x=9$ ⇒ $x=\frac{log (1)}{log (5)}=0$

Non of these results satisfy the equation.But we can use try and error method.We can see that x must be between $1.36$ and $1.11$, we try $x=1.3$:

$1.3^2=1.69=1-5^x$ ⇒ $x=\frac{log 8.31}{log 5}≈1.3$

sirous
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