Tough logarithm problem

komencanto

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2^(t/4)-2^(t/5) = 3


How does one figure this out? The answer in my maths books is wrong! I couldn't believe it after spending ages on it, I must have an answer.
 
hint: the next step is

let x = 20t

now solve for x.

edited to add: you really don't need to do that, but it makes the subsequent steps more obvious.
 
What does 20 have to do with 1/5 and 1/4?

(I'm assuming this is a homework problem, and thus I am not just going to give you the answer.)
 
Okay, now just do the substitution in the original equation and see if the next steps aren't clearer. (I just wanted to get rid of the fractions to make the problem clearer).
 
I'm really not with you. I have an exam tomorrow and although I'm almost sure this won't be in it, I'm trying to figure it out.
 
crap. I wrote the wrong thing. I'm really sorry.

let x = t/20.

Then, 2^5x = 2^4x = 3.

Now, we know that A^bc = A^b*A^c.

And take it from here....
 
I'm trying to follow along, and that hint doesn't get me very far. So now we have

2^5x - 2^4x = 3

I can take that to

2^4x * (2^x - 1) = 3

I can then take the log-base-two of both sides, but can't get past that.

It's been 21 years since I got out of college, but I like this kind of stuff. A couple of years ago, a coworker asked me to prove that with compound interest, the interest rate times the amount of time it takes the investment to double is usually around 70, which I was able to do. This problem seems like it should be easier, but I'm stuck.
 
CurtC said:
I'm trying to follow along, and that hint doesn't get me very far. So now we have

2^5x - 2^4x = 3
Click to expand...

Remember that A^bc = A^b *A^c

Therefore, 2^5x = 2^5*2^x, etc
 
But wait 2^5x does not equal 2^5 * 2^x.

You can say that 2^(5+x) = 2^5 * 2^x, but that doesn't help here.
 
CurtC said:
But wait 2^5x does not equal 2^5 * 2^x.

You can say that 2^(5+x) = 2^5 * 2^x, but that doesn't help here.
Click to expand...

Sorry, you are right.

It is A^bc = (A^b)^c
 
OK, now what? If I substitute and let a=2^x, that gives:

a^5 - a^4 - 3 = 0

How do we solve that? I'm trying to remember L'Hospital's rule and everything else, but I can't remember how to find the roots of a fifth degree polynomial.
 
I think......

(2^t)^(1/5)-(2^t)^(1/4)=3

implies

(2^t)^((1/5)/(1/4))=3

(2^t)^(4/5)=3

2^(4t/5)=3

4t/5=log2(3)

t=(5/4)*log2(3)

Is that right?

Edit - No that's complete crap, what a moron I am!
 
bah, I thought I right, but I gave two bum hints. Sorry.
 
I ran it through Mathcad. The solution gave five roots of the equation, only one of which is a real number.

I don't think it can be simplified beyond what CurtC got. :(
 
I solved the problem by iteration.

Of all the math techniques for solving problems, iteration is about the most unpleasant.
 
wollery said:
I ran it through Mathcad. The solution gave five roots of the equation, only one of which is a real number.

I don't think it can be simplified beyond what CurtC got. :(
Click to expand...


I don't have an analytical answer, but it is about 12.4, right?
 
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