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

##### 1 Answer

Jul 16, 2016

#### Answer:

#### Explanation:

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

Algebraic method:

Expanding in terms of log x and log y,

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

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

Graphical method:

The first equation has the non-logarithmic form

The second reduces to

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]}