Notes: `5^(2x)+20*5^x-125=0` Solve the equation.

Monday, 4 April 2016

We are asked to solve $\displaystyle{{5}}^{{{2}{x}}}+{20}\cdot{{5}}^{{x}}-{125}={0}$ :

Rewrite as $\displaystyle{{\left({{5}}^{{x}}\right)}}^{{2}}+{20}\cdot{{5}}^{{x}}-{125}={0}$ and let $\displaystyle{y}={{5}}^{{x}}$ to get

$\displaystyle{{y}}^{{2}}+{20}{y}-{125}={0}$ and

(y+25)(y-5)=0 so y=-25 or y=5.

y cannot be -25 as $\displaystyle{{5}}^{{x}}>{0}$ for all real x.

If y=5 then x=1. If x=1 then 25+20(5)-125=0.

x=1 is the solution.

The graph:

We are asked to solve $\displaystyle{{5}}^{{{2}{x}}}+{20}\cdot{{5}}^{{x}}-{125}={0}$ :

Rewrite as $\displaystyle{{\left({{5}}^{{x}}\right)}}^{{2}}+{20}\cdot{{5}}^{{x}}-{125}={0}$ and let $\displaystyle{y}={{5}}^{{x}}$ to get

$\displaystyle{{y}}^{{2}}+{20}{y}-{125}={0}$ and

(y+25)(y-5)=0 so y=-25 or y=5.

y cannot be -25 as $\displaystyle{{5}}^{{x}}>{0}$ for all real x.

If y=5 then x=1. If x=1 then 25+20(5)-125=0.

x=1 is the solution.

The graph:

Notes