# Thread: Logs

1. ## Logs

Use logarithm rules to rewrite each expression in terms of $\displaystyle log_{3}2$ and $\displaystyle log_{3}5$

Here's what I did... but I think I didn't do them right... please help?

a) $\displaystyle log_3{270}$

$\displaystyle = log_3{2^{7}} + log_3{5^{3}} + 17$

b) $\displaystyle log_3{\frac{64}{125}}$

$\displaystyle = log_3{64} - log_3{125}$

$\displaystyle = log_3{2^{6}} - log_3{5^{3}}$

$\displaystyle = 6log_3{2} - 3log_3{5}$

2. Originally Posted by Macleef
Use logarithm rules to rewrite each expression in terms of $\displaystyle log_{3}2$ and $\displaystyle log_{3}5$

Here's what I did... but I think I didn't do them right... please help?

a) $\displaystyle log_3{270}$

$\displaystyle = log_3{2^{7}} + log_3{5^{3}} + 17$

b) $\displaystyle log_3{\frac{64}{125}}$

$\displaystyle = log_3{64} - log_3{125}$

$\displaystyle = log_3{2^{6}} - log_3{5^{3}}$

$\displaystyle = 6log_3{2} - 3log_3{5}$
How did you get

$\displaystyle log_3{2^{7}} + log_3{5^{3}} + 17$

in the first one?

The second one is right.

3. Originally Posted by Macleef
Use logarithm rules to rewrite each expression in terms of $\displaystyle log_{3}2$ and $\displaystyle log_{3}5$

Here's what I did... but I think I didn't do them right... please help?

a) $\displaystyle log_3{270}$

$\displaystyle = log_3{2^{7}} + log_3{5^{3}} + 17$

b) $\displaystyle log_3{\frac{64}{125}}$

$\displaystyle = log_3{64} - log_3{125}$

$\displaystyle = log_3{2^{6}} - log_3{5^{3}}$

$\displaystyle = 6log_3{2} - 3log_3{5}$
Your A) part is wrong, but B) part is right.

see,

A) $\displaystyle log_3 270$

$\displaystyle = log_3(27 \times 2 \times 5)$

$\displaystyle = log_3 27 + log_3 2 + log_3 5$

$\displaystyle = log_3 3^3 + log_3 2 + log_3 5$

can you finish it now??

4. Originally Posted by Shyam
Your A) part is wrong, but B) part is right.

see,

A) $\displaystyle log_3 270$

$\displaystyle = log_3(27 \times 2 \times 5)$

$\displaystyle = log_3 27 + log_3 2 + log_3 5$

$\displaystyle = log_3 3^3 + log_3 2 + log_3 5$

can you finish it now??
yes... I see it now. How do you know the right numbers that go to 270? Is it just trial and error or is there a method to figure it out?

5. Originally Posted by Macleef
yes... I see it now. How do you know the right numbers that go to 270? Is it just trial and error or is there a method to figure it out?
you see,

$\displaystyle 270 = 27 \times 10 = 3^3 \times 2 \times 5$

I took 27 because it will be $\displaystyle 3^3$, because base of log is 3.

Did you get it now??

6. Originally Posted by Shyam
you see,

$\displaystyle 270 = 27 \times 10 = 3^3 \times 2 \times 5$

I took 27 because it will be $\displaystyle 3^3$, because base of log is 3.

Did you get it now??
Yes I do when someone does it for me... I can't solve it myself...

for example:
$\displaystyle log_3{100}$ ... the answer is $\displaystyle 2 log_3{2} + 2 log_3{5}$ ... and I don't know how to get that on my own??

7. Originally Posted by Macleef
Yes I do when someone does it for me... I can't solve it myself...

for example:
$\displaystyle log_3{100}$ ... the answer is $\displaystyle 2 log_3{2} + 2 log_3{5}$ ... and I don't know how to get that on my own??
You need more practice of these types of questions. If you have to make in terms of log 2 and log 5, then you have to think in terms of factors of 2 and 5, and their powers;

100 = factors of 2 and 5

8. ok for that question... $\displaystyle log_3{100}$, why is it $\displaystyle 5^2$ and $\displaystyle 2^2$ which equals to only 29 and not 100?

9. Originally Posted by Macleef
ok for that question... $\displaystyle log_3{100}$, why is it $\displaystyle 5^2$ and $\displaystyle 2^2$ which equals to only 29 and not 100?
you don't have to add,

you have to multiply.(only then logs get added)

$\displaystyle 5^2\times 2^2 = 100$

do you remember log laws:

$\displaystyle \log (a\times b) = \log a + \log b$

can you finish that now??

10. oh ok I get it now! I just realized what I've been doing wrong... I've been adding instead of multiplying...