Searcher2
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It is Year 2, Chapter 1, Simplify fully (a) x^2+2x-24 over 2x^2+10x times by x^2-3x over x^2+3x-18 (I understand) I don't get (b) Given ln ((x^2+2x-24)(x^2-3x))=2+ln ((2x^2+10x)(x^2+3x-18) find x in terms of e.
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zetamcfc
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(Original post by Searcher2)
It is Year 2, Chapter 1, Simplify fully (a) x^2+2x-24 over 2x^2+10x times by x^2-3x over x^2+3x-18 (I understand) I don't get (b) Given ln ((x^2+2x-24)(x^2-3x))=2+ln ((2x^2+10x)(x^2+3x-18) find x in terms of e.
Raise e to the power of both sides and see what happens.
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(Original post by zetamcfc)
Raise e to the power of both sides and see what happens.
It removes ln from both sides
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davros
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(Original post by Searcher2)
It removes ln from both sides
Not quite - what do you think happens to the 2 on the RHS?
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(Original post by davros)
Not quite - what do you think happens to the 2 on the RHS?
Does it become 2 to the power of e or some sort of logarithm? I am not sure.
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davros
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(Original post by Searcher2)
Does it become 2 to the power of e or some sort of logarithm? I am not sure.
OK, forget about all the complicated stuff at the moment.

If you have X = Y + Z, what do you get if you raise e to the power of both sides? (Hint: remember your basic rules of indices)
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(Original post by davros)
OK, forget about all the complicated stuff at the moment.

If you have X = Y + Z, what do you get if you raise e to the power of both sides? (Hint: remember your basic rules of indices)
e^x=e^yz -I am sorry if it is wrong..you could just show me solution and I can figure out how it happened
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davros
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(Original post by Searcher2)
e^x=e^yz -I am sorry if it is wrong..you could just show me solution and I can figure out how it happened
Well we're not allowed to post solutions - against the rules of the forum - so if you're struggling I'd suggest going back and revising some of the basics or get assistance from a teacher.

To start you off, what I was getting at was;

X = Y + Z \implies e^X = e^{Y+Z} = e^Y e^Z

So with your example in mind, X is the complicated expression on the left, Y = 2 and Z is the complicated expression on the right after the + sign.

Now remember that exponentiating removes the logarithms and you should get something algebraic to work with that resembles the first part of the question.
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Searcher2
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(Original post by davros)
Well we're not allowed to post solutions - against the rules of the forum - so if you're struggling I'd suggest going back and revising some of the basics or get assistance from a teacher.

To start you off, what I was getting at was;

X = Y + Z \implies e^X = e^{Y+Z} = e^Y e^Z

So with your example in mind, X is the complicated expression on the left, Y = 2 and Z is the complicated expression on the right after the + sign.

Now remember that exponentiating removes the logarithms and you should get something algebraic to work with that resembles the first part of the question.
Ok. Can you write down the 2nd step please?
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davros
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(Original post by Searcher2)
Ok. Can you write down the 2nd step please?
Well part (b) is telling you that

ln A = 2 + ln B where A and B are the rational expressions involving x that I'm not going to type out in full. So when you exponentiate both sides you'll get e^{ln A} = e^{2 + ln B} = e^2 \cdot e^{ln B} using standard indices rules.

Now use the fact that e^u is the inverse of ln u to remove the logarithms, and you should get something that looks a lot like a rearrangement of part (a) when you've replaced A and B with the corresponding expressions involving x.
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