Question 592237
<pre>
5<sup>2x-1</sup> = 3<sup>x+4</sup>

log(5<sup>2x-1</sup>) = log(3<sup>x+4</sup>)

(2x-1)log(5) = (x+4)log(3)

Let log(5) = A
Let log(3) = B

     (2x-1)A = (x+4)B

     A(2x-1) = B(x+4)

     2Ax - A = Bx + 4B

    2Ax - Bx = A + 4B

   x(2A - B) = A + 4B

           x = {{{(A+4B)/(2A-B)}}}

           x = {{{(log((5))+4log((3)))/(2log((5))-log((3)))}}}

You can leave it like that or write it as the quotient of two logs:

           x = {{{(log((5))+log((3^4)))/(log((5^2))-log((3)))}}}
           
           x = {{{(log((5))+log((81)))/(log((25))-log((3)))}}}

           x = {{{log((5*81))/log((25/3))}}}

           x = {{{log((405))/log((25/3))}}}

Edwin</pre>