Question 552114
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I can't tell which of these you meant:

4log(4x) - 3log(5y) = 1     or   4log<sub>4</sub>(x) - 3log<sub>5</sub>(y) = 1
 log(2x) -  log(5y) = 2           log<sub>2</sub>(x) -  log<sub>5</sub>(y) = 2

I tried doing it the first way and there was no solution, 
so I think you must have meant the second way.  

4log<sub>4</sub>(x) - 3log<sub>5</sub>(y) = 1
 log<sub>2</sub>(x) -  log<sub>5</sub>(y) = 2

I will change the first term in the first equation so 
it will be a log to the base 2, like the first term in
the second equation.  Using the change of base formula 
on that term:

4log<sub>4</sub>(x) = {{{4log(2,(x))/log(2,(4))}}} = {{{4log(2,(x))/log(2,(2^2))}}} = {{{4log(2,(x))/(2log(2,(2)))}}} = {{{4log(2,(x))/2}}} = 2log<sub>2</sub>(x) 

The system of equations is now:

2log<sub>2</sub>(x) - 3log<sub>5</sub>(y) = 1
 log<sub>2</sub>(x) -  log<sub>5</sub>(y) = 2

let u = log<sub>2</sub>(x) 
let v = log<sub>5</sub>(y)

The system of equations is now:

2u - 3v = 1
 u -  v = 2

Solve that system of equations by substitution or elimination
and get (u,v) = (5,3).  I'm sure you can do that.

But we don't want u and v, we want x and y.
So we substitute back:

u = log<sub>2</sub>(x), which is equivalent to the exponential equation:

x = 2<sup>u</sup> = 2<sup>5</sup> = 32

v = log<sub>5</sub>(y)

y = 5<sup>v</sup> = 5<sup>3</sup> = 125

So the solution is

(x,y) = (32,125)

Edwin</pre>