Question

Logs - please explain how they work

Answer 1

questions pls..

Answer 2

How do you work with log?

Answer 3

I don't know how to work with logarithms.

Answer 4

Logs enable muliplication and division to be carried out by addition and subtraction.
If you look at the example below you will see that when the numbers on the left hand side are multiplied their indices are added.
\[10^{2}\times10^{4}=10^{6}\]
So if we use logs to base 10, the log of 100 becomes 2 and the log of 10,000 becomes 4.
log100 + log10,000=2+4=6. The antilog of 6 is: \[10^{6}\]=1,000,000.

Can you follow the above?

Answer 5

sort of...
so the log is sort of like the amount of times number A must be multiplied by itself to become B

I understand Powers, just not Logs.

Answer 6

antilog =) nice

Answer 7

using this analogy log is antipower =)))

Answer 8

ah. I sort of get it now...
so log is NOT like Sqrt, but it is at the same time...?

log(2)256 + log(2)128 = 8+7 = 15, therefore it = 2^15, which is 32768.

Answer 9

is that right?

Answer 10

I don't get it, which one you've written ? ?
\[\LARGE \log_2 (256)+\log_2(128)\]

\[\LARGE \log_2 (256)+\log_2(128)=8+7\]

Answer 11

You've got it :)

Answer 12

I mean which was the given...
if:
\[\LARGE \log_a(b)+\log_a(c)=\log_a(b\cdot c)\]
\[\LARGE \log_a(b)-\log_a(c)=\log_a\left(\frac bc\right)\]

Answer 13

and...\[\LARGE \log_a(b)=c \quad \quad \Longrightarrow b=a^c\]

Answer 14

\[\LARGE \log_a(a)=1 \quad \quad \quad \log_a(1)=0 \]
\[ \LARGE \quad \quad \log_a(b)=\frac{1}{\log_b(a)}\]

Answer 15

\[\huge \log_{a^b}(c)=\frac 1b \cdot \log_a(c)\]
\[\huge \log_a(b)^c=c\cdot \log_a(b)\]

Answer 16

example:


\[\LARGE 3^x=3 \quad \quad \Longleftrightarrow \quad x=1\]
but if we use logs... we have:
\[\LARGE \log(3)^x=\log(3)\]
look the rule I wrote above we have,
\[\LARGE x\cdot \log(3)=\log(3)\]
\[\LARGE x=\frac{\log(3)}{\log(3)}\]
\[\LARGE x=\log_3(3)\]
\[\LARGE x=1\]

Answer 17

anyway...