Question

Expand The Logarithms : ln[ex^3/y^2z]

I would really appreciate any explanation or help that could be given about this problem or like problem.

Answer 1

\[\ln [ex ^{3}/y ^{2}z]\]

Answer 2

ln(m/n)=ln(m)-ln(n)

Answer 3

\[e ^{x}=1+x+x ^{2}/2!+x ^{3}/3!+.......\]

Answer 4

I'm not really understanding in how you got there. Where are we getting the \[x ^{2}/2!\] from ? Also where exactly did the "ln" go ?

Answer 5

You dont need that expansion.
Just remember lne = 1.
And ln( m . n) = lnm + ln(n)
So the question becomes,
\[lne + lnx^2 - lny^2 - lnz\]

Answer 6

= 1 + 2lnx - 2lny - lnz.

Answer 7

isnt it e^(x^3)?

Answer 8

Not from the first comment.

Answer 9

I'm understanding most all of it except the "+2lnx". from the original problem above ln[ex3/y2z] it shows an exponent of 3, does that get reduced or separated to become the "2lnx" ?

Answer 10

it seems as if x^3 is in power and you are saying multiplication???????

Answer 11

Oh that 3 should be an exponent not shown as being multiplied.

Answer 12

lne+3inx-2lny-lnz

Answer 13

So basically the exponent moves to the beginning of the sequence ?

Answer 14

ya

Answer 15

Alright, I understand now. Thank you very much for the help.

Answer 16

\[\log _{b}(m ^{n})=n \log _{b}(m)\]