Question 1042658
<pre><b>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:

{{{system(x*log(2) + y*log(3) = (y-x)*log(5),  
x+y = log(2) )}}}

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:

{{{system(Ax + By = C(y-x),  
x+y = A )}}}

Simplify the first equation:

{{{Ax + By = Cy-Cx}}}

{{{Ax+Cx+By-Cy=0}}}

{{{(A+C)x+(B-C)y=0}}}

{{{system((A+C)x+(B-C)y=0,  
x+y = A )}}}

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

{{{x}}}{{{""=""}}}{{{abs(matrix(2,2,0,B-C,A,1))/abs(matrix(2,2,A+C,B-C,1,1))}}}{{{""=""}}}{{{(0-A(B-C))/((A+C)-(B-C))}}}{{{""=""}}}{{{(-AB+AC)/(A+C-B+C)}}}{{{""=""}}}{{{(AC-AB)/(A+2C-B)}}}{{{""=""}}}{{{(A(C-B))/(A+2C-B)}}}

{{{x}}}{{{""=""}}}{{{(log(2)(log(5)-log(3)))/(log(2)+2log(5)-log(3))}}} &#8776; 0.0546574428

{{{y}}}{{{""=""}}}{{{abs(matrix(2,2,A+C,0,1,A))/(matrix(1,2,SAME,DENOMINATOR))}}}{{{""=""}}}{{{(A(A+C)-0)/(A+2C-B)}}}{{{""=""}}}{{{(A(A+C))/(A+2C-B)}}}

{{{y}}}{{{""=""}}}{{{(log(2)(log(2)-log(5)))/(log(2)+2log(5)+log(3))}}} &#8776; 0.2463725529

Edwin</pre></b>