Question

If $\frac{{\log }_{2}a}{4}=\frac{{\log }_{2}b}{6}=\frac{{\log }_{2}c}{3p}$ and $a^2{b}^{2}c=1$, then find the value of $-p=\frac{20}{m}$.Find $m$?

Answer 1

$\frac{lo{g}_{2}a}{4}=\frac{lo{g}_{2}b}{6}\Rightarrow lo{g}_2\left(a^6\right)=lo{g}_2\left(b^4\right)\Rightarrow a^6=b^4$

$\frac{lo{g}_{2}a}{4}=\frac{lo{g}_{2}c}{3P}\Rightarrow lo{g}_2\left(a^{3P}\right)=lo{g}_2\left(c^4\right)\Rightarrow a^{3P}=c^4$

As $a^2{b}^{2}c=1\Rightarrow a^2{a}^3{a}^{3P/4}=a^0\Rightarrow 5+\frac{3P}{4}=0$

$\Rightarrow P=-20/3$