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Log proof help

Hi,
How do you do this question?:

Given that a>0, b>0, prove that ln(ab) = lna + lnb
You have got to know your log laws to answer all logs questions.

Log(a) + Log(b) is identical to Log(ab)
Log(a) - Log9b) is identical to Log(a/b)
Original post by Earl_Earl
You have got to know your log laws to answer all logs questions.

Log(a) + Log(b) is identical to Log(ab)
Log(a) - Log9b) is identical to Log(a/b)


Thanks for your reply :smile: yes I know that, but do you know how you prove it?
Reply 3
Avatar for Naruke
Naruke
11
Original post by Phoebus Apollo
Hi,
How do you do this question?:

Given that a>0, b>0, prove that ln(ab) = lna + lnb


urm might be wrong yolo

if you let

x=ln⁡a x = \ln a
&
y=ln⁡b y = \ln b

In terms of e, this would give:

ex=a e^x = a
ey=b e^y = b

ab=exey ab = e^xe^y
Unparseable latex formula:

ab = e^x^+^y


Unparseable latex formula:

\ln ab = \ln e^x^+^y


ln⁡ab=(x+y)ln⁡e \ln ab = (x + y )\ln e
ln⁡ab=x+y \ln ab = x + y
ln⁡ab=ln⁡a+ln⁡b \ln ab = \ln a + \ln b
(edited 7 years ago)
Original post by Naruke
urm might be wrong yolo

if you let

x=ln⁡a x = \ln a
&
y=ln⁡b y = \ln b

In terms of e, this would give:

ex=a e^x = a
ey=b e^y = b

ab=exey ab = e^xe^y
Unparseable latex formula:

ab = e^x^+^y


Unparseable latex formula:

\ln ab = \ln e^x^+^y


ln⁡ab=(x+y)ln⁡e \ln ab = (x + y )\ln e
ln⁡ab=x+y \ln ab = x + y
ln⁡ab=ln⁡a+ln⁡b \ln ab = \ln a + \ln b

Thanks, Naruke :smile:
Reply 5
Avatar for Naruke
Naruke
11
Original post by Phoebus Apollo
Thanks, Naruke :smile:


No problemo

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