Question 1054592
<pre>
{{{4^(5x+1)}}}{{{""=""}}}{{{5^(4x-1)}}}

Take logs of both sides:

{{{log(4^(5x+1))}}}{{{""=""}}}{{{log(5^(4x-1))}}} 

Move the exponents in front of the logs as multipliers:

{{{(5x+1)log((4))}}}{{{""=""}}}{{{(4x-1)log((5))}}}

To make things easier, let single letters stand for
those logs:

Let's substitute the letter A for log(4) and the letter
B for log(5), substituting

log(4) = A and log(5) = B, the equation is simpler as

{{{(5x+1)A}}}{{{""=""}}}{{{(4x-1)B}}} 

or

{{{A(5x+1)}}}{{{""=""}}}{{{B(4x-1)}}}

or

{{{5Ax+A}}}{{{""=""}}}{{{4Bx-B}}}  

Solve for x:

{{{5Ax-4Bx}}}{{{""=""}}}{{{-A-B}}}

{{{(5A-4B)x}}}{{{""=""}}}{{{-A-B}}}

Divide both sides by 5A+4B

{{{((5A-4B)x)/(5A-4B)}}}{{{""=""}}}{{{(-A-B)/(5A-4B)}}}

{{{(cross((5A-4B))x)/cross((5A-4B))}}}{{{""=""}}}{{{(-A-B)/(5A-4B)}}}

{{{x}}}{{{""=""}}}{{{(-A-B)/(5A-4B)}}}

Now get your calculator.  If you have a graphing calculator
and know how, you can store log(4) as A and log(5) as B using
the STO key, and type in {{{(-A-B)/(5A-4B)}}} and press ENTER.

Or substitute log(4) for A and log(5) for B

{{{x}}}{{{""=""}}}{{{(-log(4)-log(5))/(5log(4)-4log(5))}}}

You'll get x = -6.067672624

Edwin</pre></b>