# X=\log_{12}18 and Y=\log_{24}54. Find XY+5(X−Y)

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

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Andrew Reyes
Let $$\displaystyle{I}={\frac{{{\log{{18}}}}}{{{\log{{12}}}}}}\cdot{\frac{{{\log{{54}}}}}{{{\log{{24}}}}}}+{5}{\left({\frac{{{\log{{18}}}}}{{{\log{{12}}}}}}−{\frac{{{\log{{54}}}}}{{{\log{{24}}}}}}\right)}$$ . Also, let $$\displaystyle{\log{{3}}}={x}$$ and $$\displaystyle{\log{{2}}}={y}.$$
Then,
$$\displaystyle{I}={\frac{{{{\log{{3}}}^{{2}}\cdot}{2}}}{{{{\log{{2}}}^{{2}}\cdot}{3}}}}\cdot{\frac{{{{\log{{3}}}^{{3}}\cdot}{2}}}{{{{\log{{2}}}^{{3}}\cdot}{3}}}}+{5}{\left({\frac{{{{\log{{3}}}^{{2}}\cdot}{2}}}{{{{\log{{2}}}^{{2}}\cdot}{3}}}}-{\frac{{{{\log{{3}}}^{{3}}\cdot}{2}}}{{{{\log{{2}}}^{{3}}\cdot}{3}}}}\right)}={\frac{{{2}{x}+{y}}}{{{2}{y}+{x}}}}\cdot{\frac{{{3}{x}+{y}}}{{{3}{y}+{x}}}}+{5}{\left({\frac{{{2}{x}+{y}}}{{{2}{y}+{x}}}}-{\frac{{{3}{x}+{y}}}{{{3}{y}+{x}}}}\right)}$$
$$\displaystyle={\frac{{{6}{x}^{{2}}+{5}{x}{y}+{y}^{{2}}+{10}{x}^{{2}}+{35}{x}{y}+{15}{y}^{{2}}-{35}{x}{y}-{10}{y}^{{2}}}}{{{\left({2}{y}+{x}\right)}{\left({3}{y}+{x}\right)}}}}$$
$$\displaystyle={\frac{{{x}^{{2}}+{5}{x}{y}+{6}{y}^{{2}}}}{{{x}^{{2}}+{5}{x}{y}+{6}{y}^{{2}}}}}={1}$$
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Daniel Cormack
Note that $$\displaystyle{X}{Y}+{5}{\left({X}−{Y}\right)}={\left({X}−{5}\right)}{\left({Y}+{5}\right)}+{25}$$, so it suffices to find $$\displaystyle{\left({X}−{5}\right)}{\left({Y}+{5}\right)}$$.
$$\displaystyle{\left({X}−{5}\right)}={{\log}_{{{12}}}{\left({18}\right)}}−{5}={{\log}_{{{12}}}{\frac{{{18}}}{{{12}^{{5}}}}}}={{\log}_{{{12}}}{3}^{{−{3}}}}{2}^{{−{9}}}=−{3}{{\log}_{{{12}}}{\left({24}\right)}}.$$
$$\displaystyle{\left({Y}+{5}\right)}={{\log}_{{{24}}}{\left({54}\right)}}+{5}={{\log}_{{{24}}}{\left({54}\cdot{24}^{{5}}\right)}}={{\log}_{{{24}}}{\left({2}^{{{16}}}{3}^{{8}}\right)}}={8}{{\log}_{{{24}}}{\left({12}\right)}}.$$
Multiplying together gives $$\displaystyle−{24}{{\log}_{{{12}}}{\left({24}\right)}}{{\log}_{{{24}}}{\left({12}\right)}}=−{24}{{\log}_{{{12}}}{\left({12}\right)}}=−{24}.$$
Adding $$\displaystyle{25}$$ to this gives $$\displaystyle{1}$$, which is your answer.