1. ## a few questions about lne

Attached are a few questions I have about operations concerning ln and ln e.
Any help would be greatly appreciated.
1.
The book has:
lne^(-2x)=e^(-2x)
I thought lne^(-2x)=-2x
2.
3e^(ln8/3)=3e^(ln2) How does Ln8/3=ln2?
3.1/2(ln27-ln3)=Ln3
How do we get ln3 as an answer.

2. kid funky fried posted:

1.
The book has:
$\ln e^{-2x} = e^{-2x}$

I thought:
$\ln e^{-2x} = 2x$ ?

2.
$3e^{\frac{\ln 8 }{3}} = 3e^{\ln 2}$

How does $(\ln 8) /3)=\ln 2$ ?

3.
$\frac{1}{2}(\ln 27 - \ln 3) = \ln 3$

How do we get $\ln 3$ as an answer?

Thanks.

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Firstly, $\log_c a^b=b\log_c a$

1. $\ln e^{-2x}$
The exponent will become the coefficient
${-2x}\ln e$
As ln e = 1, it's $-2x$

2. $\frac{\ln 8 }{3}$
$\ln 8 = \ln 2^3$
This is the same situation, the exponent will become the coefficient.
$\ln 8 = \ln 2^3 = 3\ln 2$
And $\frac{3\ln 2}{3} = \ln 2$

3. $\frac{1}{2}(\ln 27 - \ln 3)$
$\frac{1}{2}(\ln 3^3 - \ln 3)$
$\frac{1}{2}(3\ln 3 - \ln 3)$
$\frac{1}{2}(2\ln 3)$
$\ln 3$

There are also other ways to find these. For example,

3. $\frac{1}{2}(\ln 27 - \ln 3)$

Rule: $\log_c a - \log_c b = \log_c \frac{a}{b}$

So, $\frac{1}{2}(\ln \frac{27}{3})$

$\frac{1}{2}(\ln 9)$

$\frac{1}{2}(\ln 3^2)$

$\frac{1}{2}(2\ln 3) = \ln 3$