Help evalutating numbers with exponents

• Dec 7th 2008, 11:14 AM
ferken
Help evalutating numbers with exponents
Hi can someone show me step by step how to evaluate these problems plz. Thanks in advance =]

• Dec 7th 2008, 11:22 AM
Moo
Hello,
Quote:

Originally Posted by ferken
Hi can someone show me step by step how to evaluate these problems plz. Thanks in advance =]

Okay, there are basically 3 rules to know :
$\displaystyle a^ba^c=a^{b+c}$
$\displaystyle (a^b)^c=(a^c)^b=a^{bc}$
$\displaystyle \frac{1}{a^b}=a^{-b}$

For the first one :

$\displaystyle \frac{32^{-\frac 15}}{(8^{\frac 53}-625^{\frac 12})^{-1}}=32^{-\frac 15} \left(8^{\frac 53}-625^{\frac 12}\right)$

Note that $\displaystyle 32=2^5$, $\displaystyle 8=2^3$, $\displaystyle 625=5^4$

Hence $\displaystyle 32^{-1/5}=(2^5)^{-1/5}=2^{-1}$
$\displaystyle 8^{5/3}=(2^3)^{5/3}=2^5$
$\displaystyle 625^{1/2}=(5^4)^{1/2}=5^2$

So your expression is now :

$\displaystyle =2^{-1} \left(2^5-5^2\right)=\frac 12 \left(32-25\right)=\frac 72$
• Dec 7th 2008, 01:05 PM
ferken
Thanks for the help, but im still kinda stuck for the second one.
Basically my problem is that i try to change it so that they all have the same base, but i am having troubles.
• Dec 7th 2008, 01:21 PM
chabmgph
Try to change them into common bases, in this case, 2 and 3. And then use the rules:

Quote:

Originally Posted by Moo
Hello,

Okay, there are basically 3 rules to know :
$\displaystyle a^ba^c=a^{b+c}$
$\displaystyle (a^b)^c=(a^c)^b=a^{bc}$
$\displaystyle \frac{1}{a^b}=a^{-b}$

and also $\displaystyle (ab)^c=(a^c)(b^c)$
to combine liked terms.

You may want to try something like this:

$\displaystyle \frac{(27^{\frac14})(3^{\frac12})(6^{\frac34})}{2^ {-\frac14}}$
$\displaystyle =\frac{(3^{\frac34})(3^{\frac12})(2^{\frac34})(3^{ \frac34})}{2^{-\frac14}}$

and so forth.