How would I set up a equation to find the sum of...
4
E log24^n
n=1
It all depends on what you are actualy asking.Originally Posted by Stuart
Interpretation 1:
$\displaystyle
\sum_{n=1}^4 \log(24^n)=\log(24)+\log(24^2)+\log(24^3)+\log(24^ 4)$$\displaystyle =\log(24^{10})=10\log(24)
$
Interpretation 2:
$\displaystyle
\sum_{n=1}^4 \log_2(4^n)=\sum_{n=1}^4 \log_2(2^{2n})=\sum_{n=1}^4 2n\log_2(2)=\sum_{n=1}^4 2n=20
$
RonL
I'm afraid it does not. There is a common abuse of notation withOriginally Posted by ThePerfectHacker
trig functions which might lead you to think this, but that is the
aberrant usage not this:
$\displaystyle
f^2 (x)=f(f(x)),\ f^3(x)=f(f(f(x))),\ \dots
$
The abuse of notation that leaves:
$\displaystyle
f^n(x)=(f(x))^n
$
is in general to be deprecated as it leaves us with no compact
notation for the iterates of functions
RonL