Natural log problem

**galanm** · March 1st 2009, 03:25 PM

I've solved this problem a couple different ways and come up with different answers and I have had a couple other people come up with different answers. What log basic am I missing?

Given that ln a = 5 and ln b = 7 evaluate the following:

ln a2 be2

**arpitagarwal82** · March 1st 2009, 03:48 PM

Is it
$\ln a^2 \ln be^2$

or

$\ln a^2 \ln (be)^2$

Assuming its $\ln a^2 \ln be^2$

= 2\ln a ( \ln b + 2\ln e)
= 2\ln a ( \ln b +2)
2 x 5 ( 7 + 2)
2 x 5 x 9
90

**galanm** · March 1st 2009, 03:51 PM

Problem is presented as:

ln a^2be^2 no spaces between variables

**arpitagarwal82** · March 1st 2009, 04:04 PM

then its
$\n a^2be^2$

$= \ln a^2 + \ln b + \ln e^2$

$= 2\ln a + \ln b + 2\ln e$

= 2 x 5 + 7 + 2 x 1
= 19

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