How do you solve 8^(x-y)=3^x and 8^y=2^(x-3)?

1 Answer
Jun 11, 2016

(2^3)^y = 2^(x - 3)

2^(3y) = 2^(x - 3)

We're now in equal bases, so we can eliminate the bases.

3y = x - 3

3y + 3 = x

8^((3y + 3) - y) = 3^(3y + 3)

8^(2y + 3) = 3^(3y + 3)

This can be solved using the property n^m = a^b -> logn^m = loga^b

log8^(2y + 3) = log3^(3y + 3)

Now, we need to simplify using the property loga^n = nloga.

(2y + 3)log8 = (3y + 3)log3

2ylog8 + 3log8 = 3ylog3 + 3log3

2ylog8 - 3ylog3 = 3log3 - 3log8

y(2log8 - 3log3) = 3log3 - 3log8

Now, simplifying using the rule log_a(n) - log_a(m) = log_a(n/m):

y = (log(27/512))/(log(64/27))

Now, substituting back into the original equation to find x:

2^(3y) = 2^(x - 3)

2^(3( (log(27/512))/(log(64/27)))) = 2^(x - 3)

3( (log(27/512))/(log(64/27))) = x - 3

(log(19683/134217728))/(log(64/27)) + 3 = x

The solution set is therefore {(log(19683/134217728))/(log(64/27)) + 3, (log(27/512))/(log(64/27))} or {-7.23; -3.41}.

Hopefully this helps!