Multiply equation (1) and equation (3)
$(z^x \, z^y)\left( \dfrac{z^x}{z^y} \right) = 100\,000(10)$
$z^{2x} = 1\,000\,000$
$z^x = 1000$
Substitute zx = 1000 to equation (2)
$1000^y = 100\,000$
$10^{3y} = 10^5$
Thus,
$3y = 5$
$y = \frac{5}{3}$
Another way to solve for y †
Divide equation (1) by equation (3)
$\dfrac{z^x \, z^y}{\dfrac{z^x}{z^y}} = \dfrac{100\,000}{10}$
$z^{2y} = 10\,000$
$z^y = 100$
Substitute y = 5/3 to zy = 100 from above
$z^{5/3} = 100$
$z = 100^{3/5}$
Substitute y = 5/3 and z = 1003/5 to equation (2)
$[ \, (100^{3/5})^x \, ]^{5/3} = 100\,000$
$(100^{3x/5})^{5/3} = 100\,000$
$100^x = 100\,000$ ‡
$10^{2x} = 10^5$
Thus,
$2x = 5$
$x = \frac{5}{2}$ → you can also use logarithm to solve for x from ‡ similar to solution for y in †.
Answer
x = 5/2
y = 5/3
z = 1003/5