How do I evaluate this expression?
z^log k
Ok..just in case though
$\displaystyle z^{\log_z(k)}=k$
This is due to the fact that $\displaystyle \log_z(u(x))$ and $\displaystyle z^{u(x)}$ are inverses
and by definition $\displaystyle f(f^{-1}(u(x))=u(x)$
This is true in this case but not always
For example consider $\displaystyle x^2$ and $\displaystyle \sqrt{x}$
Ok, another math question:
using the values $\displaystyle \log_3(5)=1.4650
$ and $\displaystyle \log_3(2)=.6309$, how would I solve $\displaystyle log_3(45)$
I get as far as $\displaystyle \log_3(5)+\log_3(9)$, but I don't know what to do next.
heck, how do I use the give value of a log with something like $\displaystyle log_3(1.2)$
Ok you were good as you were
but you must see that $\displaystyle \log_3(5)+\log_3(9)=\log_3(5)+\log_3(3^2)=$$\displaystyle \log_3(5)+2\log(3)=1.4560+2$
and for the second one
$\displaystyle \log_3(1.2)=\log_3\bigg(\frac{6}{5}\bigg)=\log_3(6 )-\log_3(5)=$$\displaystyle \log_3(3\cdot{2})-\log_3(5)=\log_3(3)+\log_3(2)-\log_3(5)=1+.6309-1.4560=.1749$
Yes exactly
So using that we get $\displaystyle \log_5(x)=\log_5(\sqrt{25})=\log_5(5)=1$
So now we see that $\displaystyle \log_5(5)=1$
There are two ways of looking at this...the direct way
$\displaystyle 5^{\log_5(x)}=5^1\Rightarrow{x=5}$
or also noting that the only number that makes $\displaystyle \log_b$ equal to one is b...so since we have $\displaystyle \log_5(x)=1$ the only value that makes this true is the base value which is 5