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 *z*^{x} = 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 *z*^{y} = 100 from above

$z^{5/3} = 100$

$z = 100^{3/5}$

Substitute *y* = 5/3 and *z* = 100^{3/5} 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* = 100^{3/5}