Algebra exercise Level 1

Exercise:

If $a={{\log }_{12}}18\,\,\And \,\,\,b={{\log }_{24}}54$ , show that $ab+5\left( a-b \right)=1$

Solution: we know that ${{\log }_{c}}d=\frac{\ln d}{\ln c}$

$a={{\log }_{24}}18=\frac{\ln 18}{\ln 12}=\frac{\ln \left( 2\times {{3}^{2}} \right)}{\ln \left( {{2}^{2}}\times 3 \right)}=\frac{\ln 2+2\ln 3}{2\ln 2+\ln 3}$

$b={{\log }_{24}}54=\frac{\ln 54}{\ln 24}=\frac{\ln \left( 2\times {{3}^{3}} \right)}{\ln \left( {{2}^{3}}\times 3 \right)}=\frac{\ln 2+3\ln 3}{3\ln 2+\ln 3}$

So $a+1=\frac{\ln 2+2\ln 3+2\ln 2+\ln 3}{2\ln 2+\ln 3}=\frac{3\ln 2+3\ln 3}{2\ln 2+\ln 3}=\frac{3\left( \ln 2+\ln 3 \right)}{2\ln 2+\ln 3}$

$a-1=\frac{\ln 2+2\ln 3-2\ln 2-\ln 3}{2\ln 2+\ln 3}=\frac{\ln 3-\ln 2}{2\ln 2+\ln 3}$ Thus $\frac{a+1}{a-1}=\frac{3\left( \ln 3+\ln 2 \right)}{\ln 3-\ln 2}$

Also $b+1=\frac{\ln 2+3\ln 3+3\ln 2+\ln 3}{3\ln 2+\ln 3}=\frac{4\ln 2+4\ln 3}{3\ln 2+\ln 3}=\frac{4\left( \ln 2+\ln 3 \right)}{3\ln 2+\ln 3}$

$b-1=\frac{\ln 2+3\ln 3-3\ln 2-\ln 3}{3\ln 2+\ln 3}=\frac{2\ln 3-2\ln 2}{3\ln 2+\ln 3}=\frac{2\left( \ln 3-\ln 2 \right)}{3\ln 2+\ln 3}$ $\Leftrightarrow \frac{b+1}{b-1}=\frac{2\left( \ln 2+\ln 3 \right)}{\ln 3-\ln 2}$

Take $t=\frac{\ln 2+\ln 3}{\ln 3-\ln 2}$ so $\frac{a+1}{a-1}=3t\Rightarrow t=\frac{\frac{a+1}{a-1}}{3}=\frac{a+1}{3\left( a-1 \right)}$

Hence $\frac{b+1}{b-1}=2t=2\frac{a+1}{3\left( a-1 \right)}$ $\Leftrightarrow 3\left( a-1 \right)\left( b+1 \right)=2\left( a+1 \right)\left( b-1 \right)$

$\Leftrightarrow 3ab+3a-3b-3=2ab-2a+2b-2$ $\Leftrightarrow ab+5a-5b=1\Leftrightarrow ab+5\left( a-b \right)=1$