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halfblood007
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#1
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#1
Can someone show me the method to finding x please?
I know that x=4 but I don't know how it's found.
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spex
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#2
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Square both sides so that it becomes x^5 = 32^2. 32^2 is just a number so evaluate that, and then take the 5th root.
Or take the 5th root first so that it becomes x^(1/2) = 32^(1/5) and then square it.
For this you'll need to remember you exponent laws. (e^a)^b = e^ab.

Alternatively, if this is A level, use logarithms: (5/2)ln x = ln 32. >> ln x = (2/5)ln32. >> x = e^((2/5)ln32)
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danny111
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square both sides so you get x^5 = 32^2

then take 5th root so x = 32^(2/5)

and that 32^(1/5) = 2 i think is probably assumed that you know that, so

x = 2^2 = 4
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rav007
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(Original post by halfblood007)
Can someone show me the method to finding x please?
I know that x=4 but I don't know how it's found.

laws of logs
take logs of both sides

log x^5/2 = log 32

5/2 log x = log 32
log x = 2/5 log 32
2/5 log 32 is a constant which i will denote as k

so log x = k
x = 10^k

that should do it
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abcxyz6666
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(\sqrt[5]{32})^2
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sarubobo28
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I just typed in

\sqrt[\frac{5}{2}]{32}

in the calculator (as you would do with cube root)
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JayAyy
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x^5/2 = 32

x^5 = 32^2 [square both sides]

x^5 = 1024 [simplify]

x = 5th root of 1024 [fifth root]

x = 4 [answer]
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nuodai
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#8
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#8
(Original post by rav007)
laws of logs
take logs of both sides

log x^5/2 = log 32

5/2 log x = log 32
log x = 2/5 log 32
2/5 log 32 is a constant which i will denote as k

so log x = k
x = 10^k

that should do it
I really can't see why logs would be of any use here. All you have to do is notice that x = 32^{2/5} and either solve it using a calculator or with a bit of inspection.

I'd make a substitution u = \sqrt{x} to give u^5 = 32 \Rightarrow u = 2 \Rightarrow \sqrt{x} = 2 \Rightarrow \boxed{x= 4}
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rav007
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(Original post by nuodai)
I really can't see why logs would be of any use here. All you have to do is notice that x = 32^{2/5} and either solve it using a calculator or with a bit of inspection.

I'd make a substitution u = \sqrt{x} to give u^5 = 32 \Rightarrow u = 2 \Rightarrow \sqrt{x} = 2 \Rightarrow \boxed{x= 4}

yeah and to be honest i can see that aswel, but a lot of people who have asked me these types of questions dont see it. because this type of question is usually related with the logs on the paper. so its just what im used to telling people now. you will be surprised how many people wont even spot the half power n think to square it through.
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Josie woodward
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#10
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It's a Non calculator paper!!😭😭😭
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DFranklin
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#11
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(Original post by Josie woodward)
It's a Non calculator paper!!😭😭😭
You are replying to an 8 year old thread...
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