Question 1073878
<pre>
{{{matrix(1,3,
log((25))+log((4))-log((2))-log((5)),""="",1)}}}

Group the first two and the last two (taking out a -
in the last two changes the - signs inside the parentheses
to + :

{{{matrix(1,3,
( log((25))+log((4)) )-( log((2))+log((5)) ),""="",1)}}}

Use the principle that the SUM of two logs of numbers is 
the log of their PRODUCT:


{{{matrix(1,3,
log((25*4))-log((2*5)),""="",1)}}}

{{{matrix(1,3,
log((25*4))-log((2*5)),""="",1))}}}

Use the principle that the DIFFERENCE of two logs of numbers is 
the log of their QUOTIENT: 

{{{matrix(1,3,
log(((25*4)/(2*5))),""="",1)}}}

The 2 on the bottom cancels into the 4 on the top and gives 2 on top.
The 5 on the bottom cancels into the 25 on the top and gives 5 on top.

{{{matrix(1,3,
log(((cross(25)^5*cross(4)^2)/(cross(2)*cross(5)))),""="",1)}}}

{{{matrix(1,3,
log((5*2)),""="",1)}}}

{{{matrix(1,3,
log((10)),""="",1)}}}

The log of 10 in base 10 is 1.

{{{matrix(1,3,
1,""="",1)}}}

Therefore it is proved.

Edwin</pre>