 # Prove that log 2 </msup> &#x2061;<!-- ⁡ --> 5 + log 2 </msup> &# Sylvia Byrd 2022-07-07 Answered
Prove that ${\mathrm{log}}^{2}5+{\mathrm{log}}^{2}7>\mathrm{log}12$
What I tried so far:
${\mathrm{log}}^{2}5+{\mathrm{log}}^{2}7>\mathrm{log}3+\mathrm{log}4$
$\left(\mathrm{log}5+\mathrm{log}7{\right)}^{2}-2\cdot \mathrm{log}5\cdot \mathrm{log}7>\mathrm{log}3+\mathrm{log}4$
But it seems that I'm not even near the result.
Every suggestion / hint would be appreciated :)
EDIT: $\mathrm{log}$ means ${\mathrm{log}}_{10}$
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Without actually computing exact logs,...
${\mathrm{log}}_{10}5\cdot {\mathrm{log}}_{10}5+{\mathrm{log}}_{10}7\cdot {\mathrm{log}}_{10}7>{\mathrm{log}}_{10}12$
$\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}\frac{{\mathrm{log}}_{10}5}{{\mathrm{log}}_{10}7}+\frac{{\mathrm{log}}_{10}7}{{\mathrm{log}}_{10}5}>\frac{{\mathrm{log}}_{10}12}{{\mathrm{log}}_{10}5\cdot {\mathrm{log}}_{10}7}$
Now LHS $>2$ as it is the sum of a positive number ($\ne 1$) and its reciprocal. So it is sufficient to show that RHS $<2$, which is equivalent to:
${\mathrm{log}}_{10}12<2{\mathrm{log}}_{10}5\cdot {\mathrm{log}}_{10}7\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}{\mathrm{log}}_{5}12<{\mathrm{log}}_{10}49\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}3\cdot {\mathrm{log}}_{5}12<3\cdot {\mathrm{log}}_{10}49$
But ${12}^{3}=1728<{5}^{5}$, while ${49}^{3}>{10}^{5}$ shows $3\cdot {\mathrm{log}}_{5}12<5$ while $3\cdot {\mathrm{log}}_{10}49>5$

We have step-by-step solutions for your answer! pipantasi4
From ${5}^{3}=125$ and ${7}^{6}=117649$, we deduce that $\mathsf{l}\mathsf{o}\mathsf{g}\left(5\right)\ge \frac{2}{3}$ and $\mathsf{l}\mathsf{o}\mathsf{g}\left(7\right)\ge \frac{5}{6}$
From $3\left({6}^{7}\right)=839808$ and ${5}^{9}=1953125$, we deduce that $3\left({6}^{7}\right)\le {5}^{9}$ and hence ${12}^{8}\le {10}^{9}$. So $\mathsf{l}\mathsf{o}\mathsf{g}\left(12\right)\le \frac{9}{8}$.
Finally, we have
$\mathsf{l}\mathsf{o}\mathsf{g}\left(5{\right)}^{2}+\mathsf{l}\mathsf{o}\mathsf{g}\left(7{\right)}^{2}\ge \left(\frac{2}{3}{\right)}^{2}+\left(\frac{5}{6}{\right)}^{2}=\frac{41}{36}=\frac{82}{72}\ge \frac{81}{72}\ge \frac{9}{8}\ge \mathsf{l}\mathsf{o}\mathsf{g}\left(12\right)$

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