Logarithms

rowe · #1 $Old$ November 4th, 2009, 12:32 PM

We have a function such that $f(t) = 10 \cdot 2^{kt}$ and $f(\tfrac{1}{2}) = 3$ . Find $k$ .

$f(\tfrac{1}{2}) = 3$ , therefore:

$3 = 10 \cdot 2^{\tfrac{k}{2}}$

Taking the log to the base 2 on both sides:

$log_2 3 = log_2 10 + \frac{k}{2}$

Now, I know that $log_a xy = log_a x + log_a y$ , but this must mean that we had:

$log_2 3 = log_2 10 + log_2 2^{\tfrac{k}{2}}$

Can someone show me why (proof) that $log_2 2^{\tfrac{k}{2}} = \frac{k}{2}$ ?

And is it just me, or do those log renders look a bit squashed? The logs and the fractional exponent. Is there a better way to render these?

RRH · #2 $Old$ November 4th, 2009, 01:09 PM

$3 = 10 \cdot 2^{\tfrac{k}{2}}$

Divide both sides by ten ... it should now look like this

$\frac{3}{10}= 2^{kt}$

Now

$log 3 - log 10 = \frac{k} {2} log 2$

multiply both sides by 2

$2(log 3 - log 10) = k log 2$

Divide both sides by log 2 you have now isolated k

Sorry about this being sloppy I am not very good at using Latex

rowe · #3 $Old$ November 4th, 2009, 01:59 PM

Thanks for the solution, but I am interested in finding out why first:

$log_2 2^{\tfrac{k}{2}} = \frac{k}{2}$

I'm not familiar with the proof or general rule for why this is.

e^(i*pi) · #4 $Old$ November 4th, 2009, 02:05 PM

Quote:

Originally Posted by rowe $View Post$

Thanks for the solution, but I am interested in finding out why first:

$log_2 2^{\tfrac{k}{2}} = \frac{k}{2}$

I'm not familiar with the proof or general rule for why this is.

If you're happy with $log_c(ab) = log_c(a)+log_c(b)$ then that can be used (although I'm sure someone can make it more elegant

$x^n = x \cdot x \cdot x ...$ up to n times

$log_2(x \cdot x \cdot x ... \cdot x)$

$= log_2(x) + log_2(x) + log_2(x) + ... + log_2(x)$

Factor out like terms (and there are n of them)

$n(log_2(x))$

Wikipedia defines it in terms of the inverse (exponents): http://en.wikipedia.org/wiki/List_of...ler_operations

RRH · #5 $Old$ November 4th, 2009, 03:18 PM

Here is a website with the rules for logarithms it may be helpfull

RULES OF LOGARITHMS