# Exponents

• Jan 25th 2009, 02:32 PM
Gomaz23
Exponents
(2t^3/5t^-4)^-2 I need help in this problem

It asks to simplify the exponent but the catch is the answer must be positive exponents ant help will be appreciated greatly
• Jan 25th 2009, 02:36 PM
metlx
Quote:

Originally Posted by Gomaz23
(2t^3/5t^-4)^-2 I need help in this problem

It asks to simplify the exponent but the catch is the answer must be positive exponents ant help will be appreciated greatly

$\displaystyle (\frac {2t^3}{5t^{-4}})^{-2} = (\frac{2t^7}{5})^{-2} = \frac {5^2}{2^2t^{14}} = \frac {25}{4t^{14}}$
• Jan 25th 2009, 02:39 PM
Gomaz23
Can you explain further. thnks tho
• Jan 25th 2009, 02:40 PM
metlx
what part is troubling you?

$\displaystyle t^{-2} = \frac{1}{t^2}$

do you understand now?
• Jan 25th 2009, 02:45 PM
Gomaz23
howd you get 2sqrd
• Jan 25th 2009, 02:51 PM
masters
Quote:

Originally Posted by Gomaz23
(2t^3/5t^-4)^-2 I need help in this problem

It asks to simplify the exponent but the catch is the answer must be positive exponents ant help will be appreciated greatly

Hi Gomaz23,

Here's another approach

$\displaystyle \left(\frac{2t^3}{5t^{-4}}\right)^{-2}=\left(\frac{5t^{-4}}{2t^3}\right)^2=\frac{25t^{-8}}{4t^6}=\frac{25t^{-14}}{4}=\frac{25}{4t^{14}}$
• Jan 25th 2009, 02:51 PM
metlx
$\displaystyle \left( \frac{2t^7}{5} \right)^{-2} = \left( \left( \frac{2t^7}{5} \right)^{-1} \right)^2 = \left( \frac{5}{2t^7} \right)^2 = \frac{5^2}{2^2 \cdot t^{7\cdot2}}$
• Jan 25th 2009, 02:53 PM
mr fantastic
Quote:

Originally Posted by metlx
what part is troubling you?

$\displaystyle t^{-2} = \frac{1}{t^2}$

do you understand now?

Quote:

Originally Posted by Gomaz23
howd you get 2sqrd

It was posted as an example to try and help you see where the positive power has come from.

You are expected to know the rule $\displaystyle t^{-n} = \frac{1}{t^n}$. In your problem n = 14.

Note: If you expect to get additional help then you need to be specific in what you don't understand. Making some general comment that amounts to "I don't get it" is less than helpful in knowing what further help you need.

You might benefit from going back and reviewing the index laws you have learned so far.