Evaluate as a fraction in index form

**Razbaz** · January 30th 2013, 09:50 PM

Evaluate (a^2 * b^7) / (a^3 * b) as a fraction in index form when a = (2/5)^4 and b = (5/8)^3
Answer is 5^22/2^58, I've tried numerous ways to get to this but am struggling. When I think I'm just about there I always find a problem with my working.
Help would be greatly appreciated.

**Prove It** · January 30th 2013, 09:58 PM

I'd start by simplifying the entire expression first...

$\displaystyle \begin{align*} \frac{a^2 b^7}{a^3 b} &= \frac{b^6}{a} \\ &= \frac{\left[ \left( \frac{5}{8} \right) ^3 \right] ^6}{ \left( \frac{2}{5} \right)^4 } \\ &= \frac{ \left( \frac{5}{8} \right)^{18} }{ \left( \frac{2}{5} \right)^4 } \\ &= \frac{ \frac{5^{18}}{8^{18}} }{ \frac{ 2^4 }{ 5^4 } } \\ &= \frac{5^{18}}{8^{18}} \cdot \frac{5^4}{2^4} &= \frac{5^{22}}{8^{18} \cdot 2^4} \\ &= \frac{5^{22}}{ \left( 2^3 \right)^{18} \cdot 2^4 } \\ &= \frac{ 5^{22} }{ 2^{54} \cdot 2^4 } \\ &= \frac{ 5^{22} }{ 2^{58} } \end{align*}$

**Razbaz** · January 30th 2013, 10:00 PM

Originally Posted by Prove It

I'd start by simplifying the entire expression first...

$\displaystyle \begin{align*} \frac{a^2 b^7}{a^3 b} &= \frac{b^6}{a} \\ &= \frac{\left[ \left( \frac{5}{8} \right) ^3 \right] ^6}{ \left( \frac{2}{5} \right)^4 } \\ &= \frac{ \left( \frac{5}{8} \right)^{18} }{ \left( \frac{2}{5} \right)^4 } \\ &= \frac{ \frac{5^{18}}{8^{18}} }{ \frac{ 2^4 }{ 5^4 } } \\ &= \frac{5^{18}}{8^{18}} \cdot \frac{5^4}{2^4} &= \frac{5^{22}}{8^{18} \cdot 2^4} \\ &= \frac{5^{22}}{ \left( 2^3 \right)^{18} \cdot 2^4 } \\ &= \frac{ 5^{22} }{ 2^{54} \cdot 2^4 } \\ &= \frac{ 5^{22} }{ 2^{58} } \end{align*}$

Thanks, the 3rd last line is where I was having problems. It helps to have someone explain it to you sometimes.

Edit: can't believe I didn't see that the 8^18 couldn't be expressed as (2^3)^18. Must have missed it entirely

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