Question

The number of solution pairs $(x,y)$ of the simultaneous equations ${\log }_{1/3}(x+y)+{\log }_3(x-y)=2$, $2^{y^3}={512}^{x+1}$ is

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

The correct option is A $0$

By Simplifying the eqn:

$⟹{\log }_{1/3}(x+y)+{\log }_3(x-y)=2$

$⟹-{\log }_3(x+y)+{\log }_3(x-y)=2$

$⟹{\log }_3\frac{(x-y)}{(x+y)}=2$

$⟹\frac{(x-y)}{(x+y)}=3^2$

$⟹(x-y)=9(x+y)$

$⟹10y+8x=0$

$⟹5y+4x=0$ ..........$\left(i\right)$

Now, By Simplifying second eqn.

$⟹2^{y^3}={512}^{x+1}$

$⟹2^{y^3}=(2^9{)}^{x+1}$

$⟹2^{y^3}=2^{(9x+9)}$

$⟹y^3=9x+9$ .......................$\left(ii\right)$

Now , By substituting the value of $x$ from eqn $\left(i\right)$ in $\left(ii\right)$

we get,

$⟹y^3=9\left(\frac{-5}{4}y\right)+9$

$⟹y^3+\frac{45}{4}y-9=0$ .................$\left(iii\right)$

Eqn .......$\left(iii\right)$ 1 positive solution

and $\because$ inputs in log should be positive

i,e. $(x+y)$ and $(x-y)$ should be greater than zero which is not possible if $y$ is positive

$\therefore$ no solution is acceptable..