logarithms problem... help!!?

Log a (N) = x , log b (N) = y
1/Log N (a) = x , 1/log N (b) = y [ changing of base]
therefore, Log N (a) = 1/x , log N (b) =1/ y
log N (a)+log N(b)= x+y/xy
log N (ab) = x+y/xy
log ab (N) = xy/ x+y [again changing the base]...ans

Let log[base a](N) = x. In exponential form, this is
a^x = N

Let log[base b](N) = y. In exponential form, this is
b^y = N

If we take a^x = N and bring both sides to the power of (1/x), we get

(a^x)^(1/x) = N^(1/x), or
a = N^(1/x)

Similarly,
b = N^(1/y)

Let's multiply them together; that is, multiply the left hand sides and right hand sides. This gives us

ab = N^(1/x) N^(1/y)

On the right hand side, we have the same base and we are multiplying. For that reason, we can *add* the exponents.

ab = N^( (1/x) + (1/y) )

Merge the added fractions into one, to get

ab = N^( (x + y)/(xy) )

Now I'm going to bring both sides to (xy)/(x + y), effectively changing the exponent of N to 1.

(ab)^( (xy)/(x + y) ) = N

If we change this to logarithmic form, we get

log[base ab](N) = (xy) / (x + y)