1. ## Logarithms

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?

2. $

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

3. 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.

4. Originally Posted by rowe
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

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

RULES OF LOGARITHMS