Harold Kessler

2022-01-01

Andrew Reyes

Beginner2022-01-02Added 24 answers

Let $I=\frac{\mathrm{log}18}{\mathrm{log}12}\cdot \frac{\mathrm{log}54}{\mathrm{log}24}+5(\frac{\mathrm{log}18}{\mathrm{log}12}-\frac{\mathrm{log}54}{\mathrm{log}24})$ . Also, let $\mathrm{log}3=x$ and $\mathrm{log}2=y.$

Then,

$I=\frac{{\mathrm{log}3}^{2}\cdot 2}{{\mathrm{log}2}^{2}\cdot 3}\cdot \frac{{\mathrm{log}3}^{3}\cdot 2}{{\mathrm{log}2}^{3}\cdot 3}+5(\frac{{\mathrm{log}3}^{2}\cdot 2}{{\mathrm{log}2}^{2}\cdot 3}-\frac{{\mathrm{log}3}^{3}\cdot 2}{{\mathrm{log}2}^{3}\cdot 3})=\frac{2x+y}{2y+x}\cdot \frac{3x+y}{3y+x}+5(\frac{2x+y}{2y+x}-\frac{3x+y}{3y+x})$

$=\frac{6{x}^{2}+5xy+{y}^{2}+10{x}^{2}+35xy+15{y}^{2}-35xy-10{y}^{2}}{(2y+x)(3y+x)}$

$=\frac{{x}^{2}+5xy+6{y}^{2}}{{x}^{2}+5xy+6{y}^{2}}=1$

Then,

Daniel Cormack

Beginner2022-01-03Added 34 answers

Note that $XY+5(X-Y)=(X-5)(Y+5)+25$ , so it suffices to find $(X-5)(Y+5)$ .

$(X-5)={\mathrm{log}}_{12}\left(18\right)-5={\mathrm{log}}_{12}\frac{18}{{12}^{5}}={\mathrm{log}}_{12}{3}^{-3}{2}^{-9}=-3{\mathrm{log}}_{12}\left(24\right).$

$(Y+5)={\mathrm{log}}_{24}\left(54\right)+5={\mathrm{log}}_{24}(54\cdot {24}^{5})={\mathrm{log}}_{24}\left({2}^{16}{3}^{8}\right)=8{\mathrm{log}}_{24}\left(12\right).$

Multiplying together gives$-24{\mathrm{log}}_{12}\left(24\right){\mathrm{log}}_{24}\left(12\right)=-24{\mathrm{log}}_{12}\left(12\right)=-24.$

Adding$25$ to this gives $1$ , which is your answer.

Multiplying together gives

Adding

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