Logarithms and exponential functions Watch

Shas72
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Solve the equation x^2.5=20x^1.25, giving your ans in exact form
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Pangol
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(Original post by Shas72)
Solve the equation x^2.5=20x^1.25, giving your ans in exact form
Start by using normal index laws to get x^(something) = 20, then take logarithms of both sides (any base will do). Is there a logarithm law that can help you rewrite what you have so that there are no powers of x left?
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Shas72
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(Original post by Pangol)
Start by using normal index laws to get x^(something) = 20, then take logarithms of both sides (any base will do). Is there a logarithm law that can help you rewrite what you have so that there are no powers of x left?
I tried doing 2.5log x= log 20+ 1.25 log x.
But I get a diff ans
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Pangol
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(Original post by Shas72)
I tried doing 2.5log x= log 20+ 1.25 log x.
But I get a diff ans
What do you do after that? What is your answer and the answer you are expecting?
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ghostwalker
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(Original post by Shas72)
I tried doing 2.5log x= log 20+ 1.25 log x.
But I get a diff ans
What did you get and what is it diff to?

Also care needs to be taken using logs, as they're only valid for x>0
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Shas72
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(Original post by Pangol)
What do you do after that? What is your answer and the answer you are expecting?
2.5log x- 1.25 log x=log 20
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Shas72
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(Original post by ghostwalker)
What did you get and what is it diff to?

Also care needs to be taken using logs, as they're only valid for x>0
So I did 2.5 log x -1.25 log x =log 20
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Pangol
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(Original post by Shas72)
2.5log x- 1.25 log x=log 20
Yes, all good so far. Now what? How many lots of log x have you got altogether?
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Shas72
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(Original post by Pangol)
Yes, all good so far. Now what? How many lots of log x have you got altogether?
So I take log x (1.25)= log 20
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Pangol
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(Original post by Shas72)
So I take log x (1.25)= log 20
I think so - if you mean 1.25 log x = log 20, then I agree. (It looks slightly odd having the number at the end - it's like saying x2 instead of 2x.)

What next? You still need to end up with x = ... .
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Shas72
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#11
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(Original post by Shas72)
So I take log x (1.25)= log 20
The textbook ans is 0, and 20^0.8
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Shas72
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(Original post by Pangol)
I think so - if you mean 1.25 log x = log 20, then I agree. (It looks slightly odd having the number at the end - it's like saying x2 instead of 2x.)

What next? You still need to end up with x = ... .
X=log 20/1.25. But its wrong
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Shas72
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(Original post by Shas72)
X=log 20/1.25. But its wrong
(Original post by Pangol)
I think so - if you mean 1.25 log x = log 20, then I agree. (It looks slightly odd having the number at the end - it's like saying x2 instead of 2x.)

What next? You still need to end up with x = ... .
So log x^1.25 = log 20
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Pangol
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(Original post by Shas72)
X=log 20/1.25. But its wrong
Ah, but it is not x that is equal to that - it is log x. You need to do one more step to get x = ... .

But on reflection, it would be quicker to do this without logarithms at all. Starting with x^2.5=20x^1.25 , we can immediately get x^1.25=20, which I think is better written as x^(5/4) = 20. From this, you can go straight to one of their answers.

(Have to say that I would probably have missed x = 0 as an answer myself.)
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Pangol
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(Original post by Shas72)
So log x^1.25 = log 20
No - you have got as far as 1.25 x log(x) = log(20). The 1.25 is not inside any logarithm - it is telling you how many lots of log(x) you have got.
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old_engineer
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(Original post by Shas72)
Solve the equation x^2.5=20x^1.25, giving your ans in exact form
You could try treating the given expression as a quadratic in x^1.25
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Pangol
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(Original post by old_engineer)
You could try treating the given expression as a quadratic in x^1.25
This would have been a far better approach. And it would be far less likely that one solution would be overlooked (although there's no excuse for that really, I should have cautioned against leaping straight in with logarithms).
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Shas72
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(Original post by Pangol)
This would have been a far better approach. And it would be far less likely that one solution would be overlooked (although there's no excuse for that really, I should have cautioned against leaping straight in with logarithms).
So if I do quadratic it would be log x^5/2-log x^5/4-log 20= 0
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Shas72
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(Original post by Pangol)
This would have been a far better approach. And it would be far less likely that one solution would be overlooked (although there's no excuse for that really, I should have cautioned against leaping straight in with logarithms).
So I use substitution let u=x^5/2
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Pangol
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(Original post by Shas72)
So if I do quadratic it would be log x^5/2-log x^5/4-log 20= 0
The idea of treating it like a quadratic is that you don't need logs at all. If you put u = x^1.25, then u^2 = x^2.5. What does this turn the original equation into?

(You don't have to do this to solve it without logs, and I think that using them has made this look harder than it is.)
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