# Thread: Quick algebra question (logs)

1. ## Quick algebra question (logs)

How do I evaluate this expression?
z^log k

2. Originally Posted by reynardin
How do I evaluate this expression?
z^log k
Is that $z^{\log_{10}(k)}$?

If so The only way I can see that you can simplify this.

$z^{log_{10}(k)}=z^{\frac{\log_z(k)}{log_z(10)}}=\b igg(z^{\log_z(k)}\bigg)^{\log_z(10)}=k^{\log_z(10) }$

I cannot really see much else

3. ah, i figured out why i was having trouble. it was z^log base z of k, with the z in microprint..

4. Originally Posted by reynardin
ah, i figured out why i was having trouble. it was z^log base z of k, with the z in microprint..
Ok..just in case though

$z^{\log_z(k)}=k$

This is due to the fact that $\log_z(u(x))$ and $z^{u(x)}$ are inverses

and by definition $f(f^{-1}(u(x))=u(x)$

This is true in this case but not always

For example consider $x^2$ and $\sqrt{x}$

5. Ok, another math question:
using the values $\log_3(5)=1.4650
$
and $\log_3(2)=.6309$, how would I solve $log_3(45)$

I get as far as $\log_3(5)+\log_3(9)$, but I don't know what to do next.

heck, how do I use the give value of a log with something like $log_3(1.2)$

6. Originally Posted by reynardin
Ok, another math question:
using the values $\log_3(5)=1.4650
$
and $\log_3(2)=.6309$, how would I solve $log_3(45)$

I get as far as $\log_3(5)+\log_3(9)$, but I don't know what to do next.

heck, how do I use the give value of a log with something like $log_3(1.2)$
Ok you were good as you were

but you must see that $\log_3(5)+\log_3(9)=\log_3(5)+\log_3(3^2)=$ $\log_3(5)+2\log(3)=1.4560+2$

and for the second one

$\log_3(1.2)=\log_3\bigg(\frac{6}{5}\bigg)=\log_3(6 )-\log_3(5)=$ $\log_3(3\cdot{2})-\log_3(5)=\log_3(3)+\log_3(2)-\log_3(5)=1+.6309-1.4560=.1749$

7. thanks for being such a huge help =) I'm actually starting to get this stuff now. Still got a bit to go though... about 9 problems left that I don't understand. Can I hit them off you?

8. Originally Posted by reynardin
thanks for being such a huge help =) I'm actually starting to get this stuff now. Still got a bit to go though... about 9 problems left that I don't understand. Can I hit them off you?
Of course...I wouldnt come on here if I wasnt in the mood to be "hit off of"

9. thanks!
ok,
$1/6=6^{n+4}$
I know the answer is n=-5, but I don't know how to do the intermediate steps.

10. Originally Posted by reynardin
thanks!
ok,
$1/6=6^{n+4}$
I know the answer is n=-5, but I don't know how to do the intermediate steps.
The key to these is get the bases the same

So rewriting we get $\frac{1}{6}=6^{-1}=6^{n+4}$

taking the $\log_6$ of both sides we get

$-1=n+5\Rightarrow{n=-5}$

11. Ah.. i get it now. 1 down, 8 to go =)

$\log_5(X)=\bigg(\frac{1}{2}\bigg)\log_5(25)$

If I understand correctly, should that turn into
$\log_5(X)=\log_5(25^\frac{1}{2})$?

12. Originally Posted by reynardin
Ah.. i get it now. 1 down, 8 to go =)

$\log_5(X)=\bigg(\frac{1}{2}\bigg)\log_5(25)$

If I understand correctly, should that turn into
$\log_5(X)=\log_5(25^\frac{1}{2})$?
Yes exactly

So using that we get $\log_5(x)=\log_5(\sqrt{25})=\log_5(5)=1$

So now we see that $\log_5(5)=1$

There are two ways of looking at this...the direct way

$5^{\log_5(x)}=5^1\Rightarrow{x=5}$

or also noting that the only number that makes $\log_b$ equal to one is b...so since we have $\log_5(x)=1$ the only value that makes this true is the base value which is 5

13. yay!

Ok, for this one though, I don't really know where to start:
$\log_2(2X+6)-\log_2(X)=3$

Just give me enough to get me started, I don't want the whole problem done for me =)

14. Originally Posted by reynardin
yay!

Ok, for this one though, I don't really know where to start:
$\log_2(2X+6)-\log_2(X)=3$

Just give me enough to get me started, I don't want the whole problem done for me =)
Good! That is the spirit!

Consider $\log_2(2x+6)-\log_2(x)=\log_2\bigg(\frac{2x+6}{x}\bigg)$

15. hm, from there I get to $2^2 = \frac{6}{x}$ Good so far?

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