Question 560326
<pre>
7<sup>x - 2</sup> = 5<sup>x</sup>

Take logarithms of both sides:

log(7<sup>x - 2</sup>) = log(5<sup>x</sup>)

Use a rule of lagarithms that allows an exponent to come 
out in front as a multiplier

(x - 2)log(7) = x·log(5)

To make the algebra less bulky and cumbersome, 
let A = log(7) and B = log(5), then we have

(x - 2)A = x·B

A(x - 2) = Bx

Ax - 2A = Bx

Get both x terms on the left side 
and the non-x term on the right side:

Ax - Bx = 2A

Factor out x

x(A - B) = 2A

Divide both sides by (A - B)

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

{{{(x(cross(A-B)))/(cross(A-B))}}} = {{{2A/(A-B)}}}

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

Substitute log(7) for A and log(5) for B

x = {{{(2log(7))/(log(7)-log(5))}}}

x = 11.56654212 round to 11.5665

Edwin</pre>