SOLUTION: How to Solve (2^x)(3^y)=5^(y-x), 10^(x+y)=2

How to Solve 2^x . 3^y=5^y-x,10^x+y=2??
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~


 = ,    (1)
 =         (2)

is equivalent to

x*ln(2) + y*ln(3) = (y-x)*ln(5),    (1')
(x+y)*ln(10) = ln(2).               (2')

It is a system of two linear equations in two unknowns x and y.

Solve it by any method (Substitution, Elimination, Determinant).


 

You may have gotten as far as she did.  That system
of equations is a monster!  Since there was a power
of 10 involved it would be better to have used common 
logs (base 10) instead of natural logs.  Then we would 
have had:



To make things easier, we let those logarithms 
equal to single letters:

Let log2 = A
Let log3 = B
Let log5 = C

Then the system is simpler to work with than
carrying all that "logarithm baggage" around:



Simplify the first equation:









Then we can solve that by Cramer's rule (determinants):



 ≈ 0.0546574428



 ≈ 0.2463725529

Edwin