Question 922516
{{{4^(1-x)=3^(2x-5)}}}

We take the log of both sides:

{{{log(4^(1-x))=log(3^(2x-5))}}}

There is a rule of logarithms that allows us to move an exponent
in front as a multiplier of the log:

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

To make things easier let log(4)=A and log(3)=B

{{{(1-x)A=(2x-5)B}}}

{{{A(1-x)=B(2x-5)}}}

{{{A-Ax=2Bx-5B}}}

{{{A+5B=2Bx+Ax}}}

{{{A+5B=(2B+A)x}}}

Divide both sides of the equation by (2B+A)

{{{(A+5B)/(2B+A)=x}}}

Now replace A and B by the logs they represent:

{{{(log(4)+5*log(3))/(2*log(3)+log4)=x}}}

With a calculator we get

1.919720789 = x

Edwin</pre>