solve this exponetial equation: (1/25)^(-x^2+5x-8)a^g(x) if a

Sunday, April 13, 2014

solve this exponetial equation: (1/25)^(-x^2+5x-8)a^g(x) if a

indeed:

class="AM"> $\displaystyle{{\left(\frac{{1}}{{25}}\right)}}^{{-{{x}}^{{2}}+{5}{x}-{8}}}<{{\left(\frac{{1}}{{25}}\right)}}^{{-{2}}}$
(1)

Your friend, according me, did trick
himself.

Indeed, even if (1)
implies:

$\displaystyle-{2}<$
-x^2+5x-8 for class="AM"> $\displaystyle{\left(\frac{{1}}{{25}}\right)}<{1}$

that is:
$\displaystyle-{{x}}^{{2}}+{5}{x}-{6}>{0}$

On
the other side you can write (1) as:

class="AM"> $\displaystyle\frac{{1}}{{{25}}^{{-{{x}}^{{2}}+{5}{x}-{8}}}}<$
1/25^-2

class="AM"> $\displaystyle\frac{{1}}{{{25}}^{{-{\left({{x}}^{{2}}-{5}{x}+{8}\right)}}}}<$
1/25^(-2)

class="AM"> $\displaystyle{{25}}^{{-{\left(-{\left({{x}}^{{2}}-{5}{x}+{8}\right)}\right)}}}<{{25}}^{{-{\left(-{2}\right)}}}$

class="AM"> $\displaystyle{{25}}^{{{{x}}^{{2}}-{5}{x}+{8}}}<{{25}}^{{2}}$

class="AM"> $\displaystyle{{x}}^{{2}}-{5}{x}+{8}<{2}$

class="AM"> $\displaystyle{{x}}^{{2}}-{5}{x}+{6}<{0}$

that is
:

class="AM"> $\displaystyle-{{x}}^{{2}}+{5}{x}-{6}>{0}$

So:
WHATEVER WAY YOU CHOICE THE RESLULT IS TO BE THE
SAME!

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Help Notes

Sunday, April 13, 2014