Question 108433
If it seems right, it should solve the equation. 
Plug the value in the equation and check the result. 
{{{5^(x-2)=4^(x-1)}}}
{{{5^(log(4)-2)=4^(log(4)-1)}}}
{{{5^(.6021-2)=4^(.6021-1)}}}
{{{5^(-1.3979)=4^(-0.3979)}}}
{{{.10542=.57602}}}
Not a true statement. Not a solution.
Let's look at the problem.
{{{5^(x-2)=4^(x-1)}}}
Let's use some variables now.
Let {{{A=5^(x-2)}}} and let {{{B=4^(x-1)}}}
{{{5^(x-2)=4^(x-1)}}}
{{{A=B}}}
{{{log(5,A)=log(5,B)}}}
Also, you know from your log rules that
{{{log(5,A)=log(4,B)/log(4,5)}}}
If you put that all together you get,
{{{log(5,A)=log(5,B)}}}
{{{log(5,A)=log(4,B)/log(4,5)}}}
{{{log(4,5)*log(5,A)=log(4,B)}}}
Now back to A and B.
{{{A=5^(x-2)}}}
{{{log(5,A)=x-2}}}
{{{B=4^(x-1)}}}
{{{log(4,B)=x-1}}}
Substituting,
{{{log(4,5)*log(5,A)=log(4,B)}}}
{{{log(4,5)*(x-2)=(x-1)}}}
Let's define a constant, c,
{{{c=log(4,5)=1.161}}}
{{{c(x-2)=(x-1)}}}
{{{cx-2c=x-1}}}
{{{cx-x=2c-1}}}
{{{x(c-1)=2c-1}}}
{{{x=(2c-1)/(c-1)}}}
Now substituting for the real values,
{{{x=(2(1.161)-1)/(1.161-1)}}}
{{{x=8.213}}}
Check your answer.
{{{5^(x-2)=4^(x-1)}}}
{{{5^(8.213-2)=4^(8.213-1)}}}
{{{5^(6.213)=4^(7.213)}}}
{{{22014=22011}}}
Close enough from roundoff error.
x=8.213