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?
2026-04-05 21:45:18.1775425518
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Is it possible to solve an equation such as $x^2+5^x - 10 = 0$ without using graphical methods?
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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 $x=1.3$:
$1.3^2=1.69=1-5^x$ ⇒ $x=\frac{log 8.31}{log 5}≈1.3$
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 $$\left( \begin{array}{cc} n & x_n \\ 0 & -2.3616508 \\ 1 & -3.3004059 \\ 2 & -3.1642570 \\ 3 & -3.1613032 \\ 4 & -3.1613018 \end{array} \right)$$
$$\left( \begin{array}{cc} n & x_n \\ 0 & 1.6604150 \\ 1 & 1.3887413 \\ 2 & 1.3171449 \\ 3 & 1.3131011 \\ 4 & 1.3130890 \end{array} \right)$$