SOLUTION: Show that log5 xy = 2 log25 x + 2 log25 y. Hence or otherwise, find the values of x and y which satisfy the equations log5 xy = 10 and (log25 y/log25 x) 3/2

Question 881537: Show that log5 xy = 2 log25 x + 2 log25 y. Hence or otherwise, find the values of x and y which satisfy the equations log5 xy = 10 and (log25 y/log25 x) 3/2
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
you want to prove:

Log5(xy) = 2Log25(x) + 2Log25(y)

2Log25(x) is equal to Log25(x^2)
2Log25(y) is equal to Log25(y^2)

Log25(x^2) + Log25(y^2) is equal to Log25(x^2 * y^2) which is equal to Log25((xy)^2).

Your original equation becomes:

Log5(xy) = Log25((xy)^2)

set Log5(xy) equal to a.
set Log25((xy)^2) equal to b.

you get:

Log5(xy) = a
Log25((xy)^2) = b

If we can prove that a = b, then that proves that Log5(xy) = Log25((xy)^2) which proves the original equation is true.

The basic definition of logs states that:

Log5(xy) = a if and only if 5^a = xy

Likewise, the basic definition of logs states that:

Log25((xy)^2) = b if and only if 25^b = (xy)^2.

25^b is the same as (5^2)^b which makes this last equation equivalent to:

(5^2)^b = (xy)^2 which can also be expressed as:

5^(2b) = (xy)^2 which can also be expressed as:

(5^b)^2 = (xy)^2.

(5^b)^2 can only be equal to (xy)^2 if 5^b = xy.

you can prove this by just taking the square root of both sides of the equation and you will get 5^b = xy.

we now have:

5^a = xy
5^b = xy

This can only be true if a = b.

Since we have proven that a = b, we have also proven that Log5(xy) is equal to Log25((xy)^2) which proves the original equation is true.

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