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Differentiation

Just a little something I need clearing up...Finding out the value of x..

x^5/2 = 32

So I moved the exponent over, to make:

x = sprt 32^5


The answer is 4 but I obviously dont get that answer following my method - I thought that is how you deal with fractional exponents....?

I know it must be something simple I am forgetting :-(
(edited 7 years ago)
Reply 1
Original post by christinajane
Just a little something I need clearing up...Finding out the value of x..

x^5/2 = 32

So I moved the exponent over, to make:

x = sprt 2^5


The answer is 4 but I obviously dont get that answer following my method - I thought that is how you deal with fractional exponents....?

I know it must be something simple I am forgetting :-(


If you have x5/2=32x^{5/2} = 32 then this must mean that x=322/5x = 32^{2/5}. Which is not 2\sqrt{2}.

Do you remember your indices laws?

If you have 322/532^{2/5} this is (321/5)2=2221/2(32^{1/5})^2 = 2^2 \neq 2^{1/2}.
Reply 2
Original post by christinajane


x^5/2 = 32


Why don't you do it step by step?

x5/2x^{5/2} is (x)5(\sqrt{x})^{5}

So tex]x^{5/2} = (\sqrt{x})^5 = 32.

First step: move the 5 over. x=321/5=2\sqrt{x} = 32^{1/5} = 2.

Now you're left with a simple equation: x=2\sqrt{x} = 2. Hence, squaring both sides: x=4x = 4.

i.e: you need to square root four to get 2.
Original post by Zacken
If you have x5/2=32x^{5/2} = 32 then this must mean that x=322/5x = 32^{2/5}. Which is not 2\sqrt{2}.

Do you remember your indices laws?

If you have 322/532^{2/5} this is (321/5)2=2221/2(32^{1/5})^2 = 2^2 \neq 2^{1/2}.



Hello Zacken,

I thought that when an exponent crosses the = sign it becomes a sqrt

like

x^2 = 4

so

x = sprt 4

x = 2

I thought the same thing happened with fractional exponents crossing the = sign..

so I thought in this case

x^5/2 = 32

x = sprt 32^5

on so on but I know its wrong....
Original post by Zacken
Why don't you do it step by step?

x5/2x^{5/2} is (x)5(\sqrt{x})^{5}

So tex]x^{5/2} = (\sqrt{x})^5 = 32
.

First step: move the 5 over. x=321/5=2\sqrt{x} = 32^{1/5} = 2.

Now you're left with a simple equation: x=2\sqrt{x} = 2. Hence, squaring both sides: x=4x = 4.

i.e: you need to square root four to get 2.


Basically you just flip the fractional exponent when it crosses the = sign?

its these little things i keep forgetting!!
Reply 5
Original post by christinajane
Hello Zacken,

I thought that when an exponent crosses the = sign it becomes a sqrt

like

x^2 = 4

so

x = sprt 4

x = 2

I thought the same thing happened with fractional exponents crossing the = sign..

so I thought in this case

x^5/2 = 32

x = sprt 32^5

on so on but I know its wrong....


This is the rule: whenever you have an exponent crossing the = sign, you flip the the exponent over.

So if you move a 2 over, it becomes 1/2. i.e: x2=9x^2 = 9 means x=±91/2=±3x = \pm 9^{1/2} = \pm 3.

If you move 1/21/2 over, it becomes 2. i.e: x1/2=2x=22=4x^{1/2} = 2 \Rightarrow x = 2^2 = 4.

If you move x3/2=yx^{3/2} = y over, it becomes y2/3y^{2/3}.

If you move x5/2=32x^{5/2} = 32 over, it becomes 322/532^{2/5}.
Thank you!!

Indices and surds get me everytime!! Its my most looked at chapters in the C1 book and on mymaths!!
Reply 7
Original post by christinajane
Thank you!!

Indices and surds get me everytime!! Its my most looked at chapters in the C1 book and on mymaths!!


It'll come with practice. :yep:

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