# Logarithms and exponential functions Watch

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#2

(Original post by

Solve the equation x^2.5=20x^1.25, giving your ans in exact form

**Shas72**)Solve the equation x^2.5=20x^1.25, giving your ans in exact form

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(Original post by

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?

**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?

But I get a diff ans

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#5

Also care needs to be taken using logs, as they're only valid for x>0

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(Original post by

What do you do after that? What is your answer and the answer you are expecting?

**Pangol**)What do you do after that? What is your answer and the answer you are expecting?

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(Original post by

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

**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

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#8

(Original post by

2.5log x- 1.25 log x=log 20

**Shas72**)2.5log x- 1.25 log x=log 20

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(Original post by

Yes, all good so far. Now what? How many lots of log x have you got altogether?

**Pangol**)Yes, all good so far. Now what? How many lots of log x have you got altogether?

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#10

(Original post by

So I take log x (1.25)= log 20

**Shas72**)So I take log x (1.25)= log 20

What next? You still need to end up with x = ... .

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(Original post by

So I take log x (1.25)= log 20

**Shas72**)So I take log x (1.25)= log 20

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(Original post by

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 = ... .

**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 = ... .

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(Original post by

X=log 20/1.25. But its wrong

**Shas72**)X=log 20/1.25. But its wrong

**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 = ... .

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#14

(Original post by

X=log 20/1.25. But its wrong

**Shas72**)X=log 20/1.25. But its wrong

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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#15

(Original post by

So log x^1.25 = log 20

**Shas72**)So log x^1.25 = log 20

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#16

(Original post by

Solve the equation x^2.5=20x^1.25, giving your ans in exact form

**Shas72**)Solve the equation x^2.5=20x^1.25, giving your ans in exact form

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#17

(Original post by

You could try treating the given expression as a quadratic in x^1.25

**old_engineer**)You could try treating the given expression as a quadratic in x^1.25

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(Original post by

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).

**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).

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**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).

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#20

(Original post by

So if I do quadratic it would be log x^5/2-log x^5/4-log 20= 0

**Shas72**)So if I do quadratic it would be log x^5/2-log x^5/4-log 20= 0

(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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