Deduce the Sum of the Exponents

$1=2^0$

$\begin{aligned} 2^x &=& \dfrac{1}{2^y}\\ (2^x)(2^y) &=& 1\\ 2^{x+y} &=& 1\\ 2^{x+y} &=& 2^0\\ \therefore x+y &=& 0\\ \end{aligned}$

2^x=2^-y

so x is negative and y is positive

and -x = y

then x + y = 0

Anything to the negative power is going to be positive when it is made in its reciprocal form. Since here we have 2^x, keep "x" as -1. Then y=1. So we get x + y = -1 + 1 = 0

I also used the logarithm... $\begin{aligned} 2^{x} &= \frac{1}{2^{y}}\\ 2^{x} &= 1 \cdot 2^{-y}\\ \log_{2} 2^{x} &= \log_{2} 2^{-y}\\ x \cdot \log_{2} 2 &= -y \cdot \log_{2} 2\\ x \cdot 1 &= -y \cdot 1\\ x + y &= 0\\ \end{aligned}$

$2^x=\frac{1}{2^y}=2^{-y} \implies x=-y \implies x-(-y)=0 \implies x+y=\boxed{\large{0}}$

• $2^{x} = \frac{1}{2^{y}}$
• $2^{x} \cdot 2^{y}$ = 1
• $2^{x + y}$ = $2^{0}$
• $x + y = 0$

Exponent laws key

Isolate for 1, and solve

Product of powers give zero.

From the equation, we know that 2^x = 2^-y. So, it means x = -y. Then, x+y = 0

to have 2^x=1/2^y, you would have to have x be negative so when you inverse , it becomes positive. That means that x=-y, o the answer is 0

2^x=1/2^y=2^-y =)x= -y =)x+y=0

2^(-1) = 1/2^(1) :. (-1) + 1 = 0