mercedes36
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I know all the rules for Logarithms however slightly confused on one part.

3logy would be logy^3
but -3logy would not be logy^-3 as there are different rules for ^-3.

What would the answer be for -3logy using the same rule?
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Sir Cumference
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(Original post by mercedes36)
but -3logy would not be logy^-3 as there are different rules for ^-3.
Why do you think that? There are no different rules for negative powers.
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mercedes36
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(Original post by Sir Cumference)
Why do you think that? There are no different rules for negative powers.
because wouldn't logy^-3 = 1/logy^3 ?
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MrDeep
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-3log(Y) would become log(Y^-3) but this will become log[1/(Y^3)]
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mercedes36
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(Original post by MrDeep)
-3log(Y) would become log(Y^-3) but this will become log[1/(Y^3)]
Right thank you, I was wondering if that would make sense.
The whole question is :

Find y in terms of x where:

-3logy + 2log(x^2) - logx = log(cubedroute)y
And I so far got
log[1/y^3] + logx^4 - logx = logy^(1/3)

Would the first log and last log somehow cancel? Or do I multiply the first log with the second log, then divide the answer by logx? Please anyone assist.
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Sir Cumference
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(Original post by mercedes36)
because wouldn't logy^-3 = 1/logy^3 ?
When you write logy^-3, do you mean \log y^{-3} or (\log y)^{-3}.

These are two different things which could be why you're confused. Only the first one is equal to -3\log y.
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mercedes36
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(Original post by Sir Cumference)
When you write logy^-3, do you mean \log y^{-3} or (\log y)^{-3}.

These are two different things which could be why you're confused. Only the first one is equal to -3\log y.
As said above the initial question was to write y in terms of x from this equation:

-3logy + 2log(x^2) - logx = log(cubedroute)y

I wasn't sure what to do with the -3logy because further down it would get confusing. Would you be able to help me with some other steps in this question. (If you look above is what I got)
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Sir Cumference
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(Original post by mercedes36)
As said above the initial question was to write y in terms of x from this equation:

-3logy + 2log(x^2) - logx = log(cubedroute)y

I wasn't sure what to do with the -3logy because further down it would get confusing. Would you be able to help me with some other steps in this question. (If you look above is what I got)
I'm busy now but hopefully others can. If you're not getting replies then try starting a new thread.
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mqb2766
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(Original post by mercedes36)
As said above the initial question was to write y in terms of x from this equation:

-3logy + 2log(x^2) - logx = log(cubedroute)y

I wasn't sure what to do with the -3logy because further down it would get confusing. Would you be able to help me with some other steps in this question. (If you look above is what I got)
I'd put the all the powers as multipliers on individual log(x) and log(y) terms, then add or subtract to get ...
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