The idea is to get rid of the log and then try to manipulate the terms such that you can find what each of these are equal to.

log3(a) = x

So remove the log first by performing the following :

a = 3^x

(Recall that if xⁿ = y, then n = logx(y) )

Do the same for this :

log81(b) = y

So b = 81^y

Now notice the indices have base of 3.so we should change the base to 3 as well.

81 = 3 x 3 x 3 x 3 = 3⁴
So 81^y = (3⁴)^y = 3^(4 x y) = 3^(4y)
Which means b = 3^(4y)

Now divide a by b.

a/b = 3^x / 3^(4y)

= 3^(x - 4y)

(Applying the property a^m/a^n = a^(m - n)

Since a/b = 3^c,

Then 3^(x - 4y) = 3^c

Since the 2 terms are equal and both terms have the same base of 3, the exponents must be equal.

So x - 4y = c

Another way :


a/b = 3^c
So c = log3 (a/b)


log3(a) = x



log81(b) = y

log3(b) / log3(81) = y
(Recall that loga(b) = logc(b) / logc(a) )

log3(b) / log3(3⁴) = y

log3(b) = y x log3(3⁴)
(81 = 3⁴)

log3(b) = y x 4log3(3)
(Applying the property loga(b)ⁿ = n loga(b) )

log3(b) = 4y
(Applying the property loga(a) = 1)


Putting these together,


log3(a) - log3(b) = x - 4y
log3(a/b) = x - 4y

But recall from the beginning that c = log3(a/b)

So c = x - 4y

NNever say thanks,deeply appreciate ?