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Year 12s: what will be the first way you research your future options? >>

Year 12s: what will be the first way you research your future options? >>

powers

y((\frac{x}{f})^\frac{1}{q-1} +1) = z

guys to get y on its own, its obvious that you divide z by the whole part in the bracket. But what happens to the power of (x/z), and if it changes could you tell me why please

thanks

(edited 13 years ago)

Original post by Hrov

y((\frac{x}{z})^\frac{1}{q-1} +1) = z

guys to get y on its own, its obvious that you divide z by the whole part in the bracket. But what happens to the power of (x/z), and if it changes could you tell me why please

thanks

Nothing happens to it, the whole expression in the outer brackets is moved en masse to the denominator.

OP

Original post by ghostwalker

Nothing happens to it, the whole expression in the outer brackets is moved en masse to the denominator.

so the power doesnt change to

\frac{1}{1-q}

then?

If not i must have made a mistake in some earlier working as i know that the power should change (iv seen the answer) :frown:

:frown:

Original post by Hrov

y((\frac{x}{z})^\frac{1}{q-1} +1) = z

guys to get y on its own, its obvious that you divide z by the whole part in the bracket. But what happens to the power of (x/z), and if it changes could you tell me why please

thanks

I thought this would be about superpowers.

Words cannot express my dissapointment. :frown:

:frown:

Original post by Hrov

so the power doesnt change to

\frac{1}{1-q}

then?

Nope.

If not i must have made a mistake in some earlier working as i know that the power should change (iv seen the answer) :frown:

:frown:

If you wish to change the power to that you need to invert the fraction within the power, so to speak.

OP

Original post by ghostwalker

Nope.

If you wish to change the power to that you need to invert the fraction within the power, so to speak.

Ok ignore the above equations

y = \frac{k}{\frac{x}{z}^\frac{1}{q-1} +1}

the question asks me to simplify by setting

\frac{1}{1-q}

to "A"
But my dilemma is i have

\frac{1}{q-1}

not

\frac{1}{1-q}

(edited 13 years ago)

Original post by Hrov

Ok ignore the above equations

y = \frac{k}{\frac{x}{z}^\frac{1}{q-1} +1}

the question asks me to simplify by setting

\frac{1}{1-q}

to "A"
But my dilemma is i have

\frac{1}{q-1}

not

\frac{1}{1-q}

So how does

\frac{1}{q-1}

relate to

\frac{1}{1-q}

?

And when you know that you can use

a^b= (\frac{1}{a})^{-b}

OP

Original post by ghostwalker

So how does

\frac{1}{q-1}

relate to

\frac{1}{1-q}

?

And when you know that you can use

a^b= (\frac{1}{a})^{-b}

Im not sure but would you swap the x and z and then the power turns to its minus. I cant realy see how this has been simplified though :s-smilie:

:s-smilie:

y = \frac{k}{\frac{z}{x}^\frac{1}{1-q} +1}

y = \frac{k}{\frac{x}{z}^A +1}

Original post by Hrov

Im not sure but would you swap the x and z and then the power turns to its minus. I cant realy see how this has been simplified though :s-smilie:

:s-smilie:

y = \frac{k}{\frac{z}{x}^\frac{1}{1-q} +1}

y = \frac{k}{\frac{x}{z}^A +1}

Can't see what else you can do.

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