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Old June 22nd, 2008, 08:19 AM
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Default solving for a letter with expononets

Hi

I have to solve for the letter x, heres the equation:

Z=A(1+x) raised to 1/2

how do i deal with the 1/2 exponent?
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  #2  
Old June 22nd, 2008, 08:24 AM
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Hello,

Quote:
Originally Posted by maldous View Post
Hi

I have to solve for the letter x, heres the equation:

Z=A(1+x) raised to 1/2

how do i deal with the 1/2 exponent?
Use this rule :

(a^b)^c=a^{bc}

So here, b=1/2. Find c such that this equation simplifies.

For example, c=2.
This will give :

Z^2=\left([A(1+x)]^{1/2} \right)^2

Z^2=[A(1+x)]^1=A(1+x)

Can you continue ?
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  #3  
Old June 22nd, 2008, 10:49 AM
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Quote:
Originally Posted by Moo View Post
Hello,



Use this rule :

(a^b)^c=a^{bc}

So here, b=1/2. Find c such that this equation simplifies.

For example, c=2.
This will give :

Z^2=\left([A(1+x)]^{1/2} \right)^2

Z^2=[A(1+x)]^1=A(1+x)

Can you continue ?
Or is it this:

Z=A(1+x)^\frac{1}{2}

\frac{Z}{A}=\left(1+x\right)^\frac{1}{2}
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Old June 22nd, 2008, 10:55 AM
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Quote:
Originally Posted by masters View Post
Or is it this:

Z=A(1+x)^\frac{1}{2}

\frac{Z}{A}=\left(1+x\right)^\frac{1}{2}
Yeah, it could be

But the reasoning is the same
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Old June 22nd, 2008, 11:07 AM
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Yeah, it could be

But the reasoning is the same
I love the

I can just hear the "Na na Na na Na na !!
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Old June 22nd, 2008, 11:09 AM
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I love the

I can just hear the "Na na Na na Na na !!
It was the purpose of it
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