Question

If $\frac{a}{b}=\frac{c}{d}$ $=$ $\frac{e}{f}$ and $\frac{2a^4{b}^2+3a^2{c}^2-5e^{4}f}{2b^6+3b^2{d}^2-5f^5}$ $=$ ${\left(\frac{a}{b}\right)}^n$ then find the value of $n$.

Answer 1

Given, $\frac{a}{b}=\frac{c}{d}$ $=$ $\frac{e}{f}$

and $\frac{2a^4{b}^2+3a^2{c}^2-5e^{4}f}{2b^6+3b^2{d}^2-5f^5}={\left(\frac{a}{b}\right)}^n$

Let $\frac{a}{b}=k$

Putting $a=kb,c=kd,e=kf$ in numerator, we get

Therefore, $\frac{2k^4{b}^6+3k^4{b}^2{d}^2-5k^4{f}^5}{2b^6+3b^2{d}^2-5f^5}=k^4=(\frac{a}{b}{)}^4$

$\Rightarrow n=4$