Question

If $(4{)}^{x+y}=1$ and $(4{)}^{x-y}=4$, then the value of $x$ and $y$ will be respectively:

• A. $\frac{1}{2}$ and $-\frac{1}{2}$
• B. $\frac{1}{2}$ and $\frac{1}{2}$
• C. $-\frac{1}{2}$ and $-\frac{1}{2}$
• D. $-\frac{1}{2}$ and $\frac{1}{2}$

Answer 1

The correct option is A $\frac{1}{2}$ and $-\frac{1}{2}$

$4^{x+y}=1$ and $4^{x-y}=4$

Taking log on both sides,

$(x+y)\log 4=\log 1$ and $(x-y)\log 4=\log 4$

$(x+y)\log 4=0$ and $(x-y)=1$

Since log 4 is not equal to 0,

$(x+y)=0$ and $(x-y)=1$

Hence, $2x=1$

$\Rightarrow x=\frac{1}{2}$ and $y=-\frac{1}{2}$