Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

The question asks to simplify:

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

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.

share|improve this question
    
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

3 Answers 3

up vote 2 down vote accepted

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

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

share|improve this answer

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

share|improve this answer
    
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

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

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.