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

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Answer 1 · 2016-07-16T16:55:56

#x=100 and y=10#. The graph plots the solution point (100, 10)

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 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 and y = 10^1=10#.

Graphical method:

The first equation has the non-logarithmic form

#y = 10^(7/3) x^(-2/3) in Q_1#

The second reduces to #y = 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]}