Question

If $2.log\left(x^{2}y\right)=3+log\left(x\right)-log\left(y\right)$ and $(x-y)=3$ where $x>0,y>0$ then find the value of x.

• A. 3
• B. 2
• C. 5
• D. 4

Answer 1

The correct option is C 5

Consider the given equation.

$2log\left(x^{2}y\right)=3+logx-logy$

$log(x^{2}y{)}^2=3+log\frac{x}{y}$

$log\left(x^4{y}^2\right)-log\frac{x}{y}=3$

$log(x^4{y}^2\times \frac{y}{x})=3$

$log(xy{)}^3=3$

$(xy{)}^3={10}^3$...........($lo{g}_a^b=x\Rightarrow b=a^x$)

$xy=10\Rightarrow y=\frac{10}{x}$ ........ (1)

And $(x-y)=3$$........\left(2\right)$

substitute equation (1) in (2)

$x-\frac{10}{x}=3$

$\frac{x^2-10}{x}=3$

$x^2-3x-10=0$

$x^2-5x+2x-10=0$

$x(x-5)+2(x-5)=0$

$(x-5)(x+2)=0$

$x=5or-2$

But $x>0$

Hence, $x=5$ is the answer.