Question

Huck correctly solved 2(4^x) - 9^x = -6^x and arrived at the solution x = log_2/3 (h). Find the numerical value of h. Please show all work!

Answer 1

take the natural log of both sides

Answer 2

Natural log of both sides? I don't really see why since there is no e in the equation. Can you please explain?

Answer 3

I think that since all the powers have different bases, then it doesn't matter what base you use for the logarithm at first. I anticipate change-of-base operations happening here.

Answer 4

is this the question:\[2(4^x)-9^x=-6^x\]

Answer 5

Yes. THat is the problem.

Answer 6

ok, so first rephrase this in terms of prime factors to get:\[2(2^{2x})-3^{2x}=-(2*3)^x\]\[2(2^{2x})-3^{2x}=-2^x3^x\]then let:\[a=2^x\]\[b=3^x\]to get:\[2a^2-b^2=-ab\]\[2a^2-b^2+ab=0\]which can be factored to get:\[(2a-b)(a+b)=0\]can you solve from here?

Answer 7

Hmm, clever. That explains the base of 2/3 for the logarithm..

Answer 8

thx CliffSedge - it took a while to realise that was the way to go :)

Answer 9

2(2^x) - 3^x = 0
log_2 (x) - log_3 (x) = 0
Am I on the right path? I'm not very sure...

Answer 10

how did you get that step? you have two different log bases there?

Answer 11

I plugged a and b back into the first binomial.

Answer 12

we ended up with:\[(2a-b)(a+b)=0\implies 2a=b\text{, or }a=-b\]therefore:\[2(2^x)=3^x\]or:\[2^x=-3^x\qquad\text{this has no real solutions so can be ignored}\]therefore the only real solution is:\[2(2^x)=3^x\]which can be rewritten as:\[2^{x+1}=3^x\]I'll leave you to solve for x from here...

Answer 13

Alright. Thanks! Are there no solutions for the other one because there can't be a negative log?

Answer 14

correct - the other one has no real solutions.

Answer 15

Just to make sure that I got it correctly:
2 * 2^x/3^x = 1
(2/3)^x = 1/2
x = log_2/3 (1/2)
and since the question asks for h, the answer is 1/2?

Answer 16

perfect!

Answer 17

Again, thanks so much! :D

Answer 18

yw :)