Harold Kessler

2022-01-01

$X={\mathrm{log}}_{12}18$ and $Y={\mathrm{log}}_{24}54$. Find $XY+5\left(X-Y\right)$

Andrew Reyes

Expert

Let $I=\frac{\mathrm{log}18}{\mathrm{log}12}\cdot \frac{\mathrm{log}54}{\mathrm{log}24}+5\left(\frac{\mathrm{log}18}{\mathrm{log}12}-\frac{\mathrm{log}54}{\mathrm{log}24}\right)$ . Also, let $\mathrm{log}3=x$ and $\mathrm{log}2=y.$
Then,
$I=\frac{{\mathrm{log}3}^{2}\cdot 2}{{\mathrm{log}2}^{2}\cdot 3}\cdot \frac{{\mathrm{log}3}^{3}\cdot 2}{{\mathrm{log}2}^{3}\cdot 3}+5\left(\frac{{\mathrm{log}3}^{2}\cdot 2}{{\mathrm{log}2}^{2}\cdot 3}-\frac{{\mathrm{log}3}^{3}\cdot 2}{{\mathrm{log}2}^{3}\cdot 3}\right)=\frac{2x+y}{2y+x}\cdot \frac{3x+y}{3y+x}+5\left(\frac{2x+y}{2y+x}-\frac{3x+y}{3y+x}\right)$
$=\frac{6{x}^{2}+5xy+{y}^{2}+10{x}^{2}+35xy+15{y}^{2}-35xy-10{y}^{2}}{\left(2y+x\right)\left(3y+x\right)}$
$=\frac{{x}^{2}+5xy+6{y}^{2}}{{x}^{2}+5xy+6{y}^{2}}=1$

Daniel Cormack

Expert

Note that $XY+5\left(X-Y\right)=\left(X-5\right)\left(Y+5\right)+25$, so it suffices to find $\left(X-5\right)\left(Y+5\right)$.
$\left(X-5\right)={\mathrm{log}}_{12}\left(18\right)-5={\mathrm{log}}_{12}\frac{18}{{12}^{5}}={\mathrm{log}}_{12}{3}^{-3}{2}^{-9}=-3{\mathrm{log}}_{12}\left(24\right).$
$\left(Y+5\right)={\mathrm{log}}_{24}\left(54\right)+5={\mathrm{log}}_{24}\left(54\cdot {24}^{5}\right)={\mathrm{log}}_{24}\left({2}^{16}{3}^{8}\right)=8{\mathrm{log}}_{24}\left(12\right).$
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.

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