# How do you solve for x and y: log(x^2y^3)=7 and log(x/y)=1?

Jul 16, 2016

$x = 100 \mathmr{and} y = 10$. The graph plots the solution point (100, 10)

#### Explanation:

For the logarithms to be real, both x and y > 0.

Algebraic method:

Expanding in terms of log x and log y,

$2 \log x + 3 \log y = 7 \mathmr{and}$

#log x - log y =1. Solving for log x and log y,

log x =2 and log y = 1. Inverting,

$x = {10}^{2} = 100 \mathmr{and} y = {10}^{1} = 10$.

Graphical method:

The first equation has the non-logarithmic form

$y = {10}^{\frac{7}{3}} {x}^{- \frac{2}{3}} \in {Q}_{1}$

The second reduces to $y = \frac{x}{10} \in {Q}_{1}$.

Look for common point at (100, 10), in the combined graph.
graph{(y-10^(7/3)x^(-2/3))(x-10y)((x-100)^2+(y-10)^2-.01)=0[90 110 5 15]}