Question

If $lo{g}_5(3^x-4^y)=3$ and $3^{\frac{x}{2}}-2^y=5$ ,then $\frac{x}{y}$ is equal to

• A. $\frac{2\left(lo{g}_{2}5\right)-2}{1+lo{g}_{2}5}$
• B. $\frac{\left(lo{g}_{2}5\right)+2}{1+lo{g}_{2}5}$
• C. $\frac{2\left(lo{g}_{2}5\right)+2}{1+lo{g}_{2}5}$
• D. $\frac{2\left(lo{g}_{2}5\right)+1}{1+lo{g}_{2}5}$

Answer 1

The correct option is C $\frac{2\left(lo{g}_{2}5\right)+2}{1+lo{g}_{2}5}$

We have, $lo{g}_5(3^x-4^y)=3$

$\Rightarrow lo{g}_5\left[\right(3^{\frac{x}{2}}{)}^2-\left(2^y{)}^2\right]=3$

$\Rightarrow lo{g}_5\left[\right(3^{\frac{x}{2}}-2^y\left)\right(3^{\frac{x}{2}}+2^y)=3[\because a^2-b^2=(a-b)(a+b)]$

$\Rightarrow lo{g}_5\left[5\right(3^{\frac{x}{2}}+2^y)=3$[Given]

$\Rightarrow lo{g}_5\left(5\right)+lo{g}_5(3^{\frac{x}{2}}+2^y)=3[\because log(ab)=loga+logb]$

$\Rightarrow 1+lo{g}_5(3^{\frac{x}{2}}+2^y)=3[\because lo{g}_a(a)=aloga]$

$\Rightarrow lo{g}_5(3^{\frac{x}{2}}+2^y)=2$

$\Rightarrow lo{g}_5(3^{\frac{x}{2}}+2^y)=2lo{g}_5\left(5\right)[\because lo{g}_a(a)=aloga]$

$\Rightarrow lo{g}_5(3^{\frac{x}{2}}+2^y)=lo{g}_5\left(5{)}^2\right[\because blog\left(a\right)=log{a}^b]$

$\Rightarrow 3^{\frac{x}{2}}+2^y=25..\left(1\right)$

and, $3^{\frac{x}{2}}-2^y=5...\left(2\right)$

Adding $\left(1\right)$ and $\left(2\right)$, we get

$2\times 3^{\frac{x}{2}}=30$

$\Rightarrow 3^{\frac{x}{2}}=15$

$\Rightarrow lo{g}_3\left(3^{\frac{x}{2}}\right)=lo{g}_3(3\times 5)$

$\Rightarrow \frac{x}{2}lo{g}_3\left(3\right)=lo{g}_3\left(3\right)+lo{g}_3\left(5\right)$

$\Rightarrow \frac{x}{2}=1+lo{g}_3\left(5\right)$

$\Rightarrow x=2[1+lo{g}_3(5\left)\right]$

Subtracting $\left(2\right)$ from $\left(1\right)$, we get

$2\times 2^y=20$

$\Rightarrow 2^y=10$

$\Rightarrow lo{g}_2\left(2^y\right)=lo{g}_2(2\times 5)$

$\Rightarrow ylo{g}_2\left(2\right)=lo{g}_2\left(2\right)+lo{g}_2\left(5\right)$

$\Rightarrow y=1+lo{g}_2\left(5\right)$

$\therefore ,\frac{x}{y}=\frac{2[1+lo{g}_3(5\left)\right]}{1+lo{g}_2\left(5\right)}$