# Thread: Having trouble solving for n in the expontents place with uncommon base.

1. ## Having trouble solving for n in the expontents place with uncommon base.

I have the equation:

16^n < 8^(n+1)

And need to solve for n.

What I did:

8^(4/3) = 16

n < (4/3)(n+1)
n < 4/3n + 4/3
3n < 4n + 4
-n < 4
n > 4

But, in the back of the book the answer is n < 3. What did I do wrong?

2. Originally Posted by thyrgle
I have the equation:

16^n < 8^(n+1)

And need to solve for n.

What I did:

8^(4/3) = 16

n < (4/3)(n+1)
n < 4/3n + 4/3
3n < 4n + 4
-n < 4
n > 4

But, in the back of the book the answer is n < 3. What did I do wrong?
Your claim is correct: $8^{4/3} =16$. So that implies that the inequality can be written as:

$(8^{4/3})^n < 8^{n+1}\implies 8^{4n/3}<8^{n+1}$

So this is true when $\frac{4}{3}n and this will give you the conclusion that's desired ( $n<3$).

I hope this clarifies things.

3. Ok thanks, I switched them around, my brain is dead today.