klepkowy7c

2022-07-15

Solve ${5}^{2x+2}-{5}^{x+2}+6=0$
How do we solve ${5}^{2x+2}-{5}^{x+2}+6=0$? I know I have to use logarithms but I am not sure how to do it.

Carassial3

Expert

Using ${a}^{mx+n}={a}^{n}\left({a}^{x}{\right)}^{m},$
we have
$25\left({5}^{x}{\right)}^{2}-25\left({5}^{x}\right)+6=0$
which is a Quadratic Equation in ${5}^{x}$
${5}^{x}=\frac{25±5}{50}=\frac{2}{5},\frac{3}{5}$
Taking logarithm,
$x\mathrm{log}5=\mathrm{log}2-\mathrm{log}5,\mathrm{log}3-\mathrm{log}5$

amacorrit80

Expert

Let
${5}^{x+2}=y$
$\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}25×{5}^{x}=y$
$\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}{5}^{x}=\frac{y}{25}$
$\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}{5}^{2x}=\frac{{y}^{2}}{625}$
$\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}{5}^{2x+2}=\frac{{y}^{2}}{25}$
The equation then, shall reduce to
$\frac{{y}^{2}}{25}-y+6=0$
$\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}{y}^{2}-25y+150=0$
This can be solved using quadratic equations.

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