# What have I done wrong in solving this problem with indices rules?

$$\left(\dfrac{25x^4}{4}\right)^{-\frac{1}{2}}.$$

So I used $(a^m)^n=a^{mn}$ to get

$$\dfrac{25}{4}x^{-2} = \dfrac{25}{4} \times \dfrac{1}{x^2} = \frac{25}{4x^2} = \frac{25}{4}x^{-2}$$

However, this isn't the answer, and I can't see what I've done wrong.

This is what the mark scheme says:

$$\left(\dfrac{25x^4}{4}\right)^{-\frac{1}{2}} = \left[\left(\frac{4}{25x^4}\right)^{\frac{1}{2}} \text{ or } \left(\frac{5x^2}{2}\right)^{-1} \text{ or } \frac{1}{\left(\dfrac{25x^4}{4}\right)^{\frac{1}{2}}}\right] = \frac{2}{5}x^{-2}$$

To me, my answer looks more simple than theirs, and I can't see what I've done wrong.

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Also $25/4$ has the power $-1/2$. You forgot it. Also $x^{-2}=1/x^2$. –  Sigur May 11 '13 at 19:09
Also $x^{-2} = \frac{1}{x^2}$, not $\frac{x}{2}$. –  user77356 May 11 '13 at 19:10
Sorry I was copying out a question with another there and got some numbers muddled up, I've edited it to what my solution actually was. –  Olly Price May 11 '13 at 19:14
Here, you could say you have 3 a to deal with. $a_1=x^4$, $a_2 = 25$ and $a_3=\dfrac14$. –  Jerry May 11 '13 at 19:15

$$(\dfrac{25x^4}{4})^{-\dfrac{1}{2}}$$ $$(\dfrac{25}{4})^{-\dfrac{1}{2}}(x^{-2})$$ $$(\dfrac{4}{25})^{\dfrac{1}{2}}(x^{-2})$$

$$(\dfrac{2}{5})(x^{-2})$$ $$\dfrac {2}{5x^2}$$

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Just to point out the "rule" you forgot:

It is correct that $$(a^m)^n=a^{mn}$$

But you have more than $(x^4)^{-1/2}$ to consider. You need to apply the fact that when a product and/or quotient of numbers/variables, enclosed in parentheses, is raised to a power, you need to distribute that power across the product in parentheses:

$$(ab)^n = a^nb^n\quad \text{or given,} \quad \left(\frac{ab}c\right)^n = \dfrac{a^nb^n}{c^n}$$

So in your case, you need to combine the two rules: $$(ab^m)^n = a^n\cdot \left(b^{m}\right)^n = a^n \cdot b^{mn}$$

So for $$\left(\dfrac{25x^4}{4}\right)^{-\frac{1}{2}}$$ I would simplify matters to express this as $$\left(\frac{25}{4} x^4\right)^{-1/2} = \left(\frac{25}{4}\right)^{-1/2} (x^4)^{-1/2} = \left(\frac{5^2}{2^2}\right)^{-1/2} (x^4)^{-2} = \frac{5^{-1}}{2^{-1}}x^{-2} = \frac 25 x^{-2} = \frac 2{5 x^2}$$

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Thanks, @Amzoti...slowish day for me. –  amWhy May 12 '13 at 0:50
me too! However I just saw a question on the chaotic Van Der Pol oscillator and it is calling my name! :-) –  Amzoti May 12 '13 at 0:52

$$\Big(\frac{25x^4}{4}\Big)^{-\frac{1}{2}}=\Big(\frac{4}{25x^4}\Big)^{\frac{1}{2}}=\frac{4^{\frac{1}{2}}}{25^{\frac{1}{2}}(x^4)^{\frac{1}{2}}}=\frac{2}{5x^2}$$

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