Question

URGENT.
If 2^(x+4) = 3^x. Prove by contradiction that x is an irrational number.

Answer 1

Well, I think if we are trying to prove something with a contradiction, we should start by letting x be what it shouldn't be, and then searching for a logical failure...

Answer 2

I started with x being a/b which is rational form, but i dont know how to prove that its not.

Answer 3

So, if we plug that in, we get...
\(\displaystyle 2^{a/b + 4} = 3^{a/b} \)

Answer 4

yeah, so the only operations that i found that we could do would be base rooting, but that doesnt seem to help

Answer 5

We could take the logarithm of both sides here.
\( \displaystyle \log \left( 2^{a/b + 4} \right) = \log \left(3^{a/b} \right) \)
\(\displaystyle (\frac{a}{b} + 4) \log 2 = \frac{a}{b} \log 3 \)

Answer 6

Well, I think if we get a/b on one side of the equality, there should be some expression of logarithms on the other side.

Answer 7

thanks for your help btw

Answer 8

We could do that, although I think it'd be easier even like this:
\( \displaystyle \frac{a}{b} \log 2 + 4 \log 2 = \frac{a}{b} \log 3 \)
and we subtract \( \frac{a}{b} \log 3 \) from both sides and \(4\log 2 \) from both sides.
We can then factor out an a/b and divide...
\( \displaystyle \frac{a}{b} \left( \log 2 - \log 3 \right) = - 4 \log 2 \)
\( \displaystyle \frac{a}{b} = \frac{ -4 \log 2 }{ \log 2 - \log 3} \)

Answer 9

I think from there, its pretty clear that our right-hand side is not rational... -4log 2 and log2 - log3 are not integers, so their quotient would not be rational.

Answer 10

You're welcome! :)

Answer 11

hey can you do one more thing
can you just delete your messages with the answers in them?
thanks man

Answer 12

this problem isnt for anything important, but sometimes online help is looked down upon for homework