# evaluation of exponential expression

• Apr 1st 2010, 07:26 PM
dymand68
evaluation of exponential expression
without a calculator, how do i determine that :
32^3/5
= 8
• Apr 1st 2010, 07:33 PM
mr fantastic
Quote:

Originally Posted by dymand68
without a calculator, how do i determine that :
32^3/5
= 8

$(32^{1/5})^3$ and you're expected to know what $32^{1/5}$ is.
• Apr 1st 2010, 07:34 PM
tonio
Quote:

Originally Posted by dymand68
without a calculator, how do i determine that :
32^3/5
= 8

Definitions: $32^{3\slash 5} = \left(32^{1\slash 5}\right)^3=\left(\sqrt[5]{32}\right)^3$ . Now just check that $2^5 = 32$ so...

Tonio
• Apr 1st 2010, 07:54 PM
dymand68
the problem is:
evaluate:
32^-3/5
=1/32^3/5
=???
1/8 ?
thus my original question: without a calculator, how do i Know that
32^3/5
= 8
...okay, find the fifth root of 32 and cube it, gotcha
• Apr 1st 2010, 10:02 PM
integral
If you want to be fancy (Giggle)

This does not require a calculator at all!

$32^{3/5}=n$
(take the log of both sides)

$\textrm{log}_{32}32^{\frac{3}{5}}=\textrm{log}_{32 }8$

$\textrm{log}_a(a)^x=x\therefore$

$\frac{3}{5}=\textrm{log}_{32}8$
log base change formula. (changes and base to two base 10 logs):
$\textrm{log}_ab=\frac{\textrm{log}(b)}{\textrm{log }(a)} \therefore$

$\frac{3}{5}=\frac{\textrm{log}8}{\textrm{log}32}$

$\textrm{log}(a)^x=x\textrm{log}a$
and
$\textrm{log}(ab)=\textrm{log}(a)+\textrm{log}(b)$

$\frac{3}{5}=\frac{3\textrm{log}2}{\textrm{log}4+\t extrm{log}8}$

$\frac{3}{5}=\frac{3\textrm{log}2}{2\textrm{log}2+3 \textrm{log}2}$

$\frac{3}{5}=\frac{3\textrm{log}2}{5\textrm{log}2}$

$\frac{3}{5}=\frac{3}{5}$

$\therefore 32^{3/5}=8$

((This is my 100th post! Whooo! :D ))