Question

how would you do ln6-ln2 without a calculator?

Answer 1

what do you mean by "do"?

Answer 2

That would be ln3

Answer 3

*find the value

Answer 4

given logs of the same base; they act like exponents. Exponents subtract when you divide like bases:

\[\frac{b^n}{b^m}=b^{n-m}\]

\[log_B(n)-log_B(m)=log_B(\frac{n}{m})\]

Answer 5

ln6 - ln2 = ln 6/2 = ln3

Answer 6

so you just move it over? thanks this really helps!

Answer 7

yep, it "looks like" you move it over, but that is just the result of the math behind it :)

Answer 8

so on this next one, 2ln6. It's completely different?

Answer 9

correct, that is a different set up that maths up differently.

Answer 10

the product of a constant and a log is equal to the log of the argument raised to the constant as a power:
\[c\ log_B(N)=log_B(N^c)\]

Answer 11

2 ln(6) = ln(6^2) = ln(36)

Answer 12

Hmm, I see. I sort of understand what you're saying. 6ln2 would be the same, just flipped. I'm guessing.

Answer 13

yep

Answer 14

the relationship between logs and exponents is:
\[log_B(n)=m\ <->\ B^m = n\]

Answer 15

they are inverses of each other and therefore share similar properties

Answer 16

mk, ln6+ln2, ln6/2?

Answer 17

addition is the opposite of subtraction; you notice that we divide when they subtract ... when they add, what do you think would be the opposite result?

Answer 18

multiplication c:

Answer 19

yep :)

ln(6) + ln(2) = ln(6*2)

Answer 20

that's it? ln(12)

Answer 21

yep

Answer 22

yayayay

Answer 23

log1/3 would beeee a subtraction then?

Answer 24

yep

Answer 25

log1-log3.

Answer 26

Lol I think.

Answer 27

log1-log3 = log1/3 and vice versa

Answer 28

Math genius.

Answer 29

l

Answer 30

log48?