# Natural log help

• Dec 7th 2009, 02:36 PM
lsnyder
Natural log help
I am having trouble on these two problems:

If a=2, b=3 , and c=5, evaluate the following:

1. ln(((a^-3)(b^1))/(bc)^-2

2. (ln(c^1))(ln(a)/(b^4))^-1

For #1, I keep getting 15.
So I have not been able to go on to #2 because I do not understand #1.

Thanks, lsnyder
• Dec 7th 2009, 02:42 PM
e^(i*pi)
Quote:

Originally Posted by lsnyder
I am having trouble on these two problems:

If a=2, b=3 , and c=5, evaluate the following:

1. ln(((a^-3)(b^1))/(bc)^-2

2. (ln(c^1))(ln(a)/(b^4))^-1

For #1, I keep getting 15.
So I have not been able to go on to #2 because I do not understand #1.

Thanks, lsnyder

1. Move that $(bc)^{-2}$ into the numerator and make the exponent +2 then use the laws of logs to separate the log term (remember that $b^2c^2$ must be applied to both terms)

$b^2c^2 [-3ln(a)+ln(b)] = 3^25^2(-3ln(2)+ln(3)) = 225(-3ln(2)+ln(3))$

Not much point simplifying that unless you require a decimal approximation
• Dec 7th 2009, 02:55 PM
lsnyder
I keep getting that as the wrong answer.

Its suppose to be like simple addition and subtraction.

ln (a^-3) (b^1) - ln (bc)^-2
ln (a^-3) (b^1) - ln (b^-2) (c^-2)
3 ln a - ln b + 2 ln b + 2 ln c
3 ln 2 - ln 3 + 2 ln 3 + 2 ln 5
6 -3 +5 +7
15

Something like that.....
Supposedly
:confused:
• Dec 7th 2009, 03:26 PM
e^(i*pi)
Quote:

Originally Posted by lsnyder
I keep getting that as the wrong answer.

Its suppose to be like simple addition and subtraction.

ln (a^-3) (b^1) - ln (bc)^-2
ln (a^-3) (b^1) - ln (b^-2) (c^-2)
3 ln a - ln b + 2 ln b + 2 ln c
3 ln 2 - ln 3 + 2 ln 3 + 2 ln 5
6 -3 +5 +7
15

Something like that.....
Supposedly
:confused:

Logarithms do not work that way, only powers of e will have rational natural logs

Did you mean $ln \left(\frac{a^{-3}b^1}{(bc)^{-2}}\right)$

If so simplify inside first: $\frac{a^{-3}b^1}{(bc)^{-2}} = \frac{b^3c^2}{a^3}$

$ln \left(\frac{b^3c^2}{a^3}\right) = ln(b^3) +ln(c^2) - ln(a^3) = 3ln(b)+2ln(c)-3ln(a)$

Sub in for a,b and c. That cannot be simplified more and any decimal answer will be an approximation