A Very Simple Problem

We are supposed to find the value of $z$ , which is equal to ${ x }^{ y }$ . To do this we have to find the values of $x$ and $y$ .

First, let us set up the equation ${ x }^{ 2 }+{ y }^{ 2 }+{ x }^{ 2 }-{ y }^{ 2 }=25+7$ . This simplifies to $2{ x }^{ 2 }=32$ , then ${ x }^{ 2 }=16$ , so $x=4$ .

Then let us solve for the equation ${ 4 }^{ 2 }+{ y }^{ 2 }=25$ . We get $16+{ y }^{ 2 }=25$ , then ${ y }^{ 2 }=9$ , and finally $y=3$ .

Finally, let us plug 4 and 3 into the equation z= ${ x }^{ y }$ . We get $z={ 4 }^{ 3 }=4 \times 4 \times 4=64$ . So $z=64$ .