Question

Calculus II - Logarithms

Express the given quantity as a single logarithm.

(1/5) ln (x+2)^(5) + (1/2) [ln(x)-ln(x^(2)+3x+2)^(2)]

Answer 1

\[\frac{1}{5}\ln(x+2)^5 +\frac{1}{2}[\ln(x)-\ln(x^2+3x+2)^2]\]natural logarithms work the same way as log functions, same rules, so \[\ln(x^2) \iff 2\ln(x)\]\[\ln(a) +\ln(b) = \ln(a \cdot b)\]

Answer 2

And always remember what your main function is, that would be the addition.

Answer 3

\[\frac{1}{5}\ln(x+2)^5 \implies \left[\ln(x+2)^5\right]^{1/5} \implies ~?\]

Answer 4

5th root ( ln ((x+2)^(5)) ?

Answer 5

What is \(\large (a^5)^{1/5} = ~?\)

Answer 6

a

Answer 7

Same rule applies to this:

\(\color{blue}{\text{Originally Posted by}}\) @Jhannybean
\[\frac{1}{5}\ln(x+2)^5 \implies \left[\ln(x+2)^5\right]^{1/5} \implies ~?\]
\(\color{blue}{\text{End of Quote}}\)

Think of \(\ln(x+2)\) as \(a\).

Answer 8

Is it just ln(x+2) then?

Answer 9

Yep \(\checkmark\)

Answer 10

Now for \(\frac{1}{2}[\ln(x) -\ln(x^2+3x+2)^2]\) We could distribute the \(\frac{1}{2} t both terms inside the brackets. What would we get?

Answer 11

\(\frac{1}{2}\)*

Answer 12

(1/2) ln (x) - (1/2) ln ((x^(2) + 3x + 2)^(2)) =
ln ( sqrt (x)) - ln ((x^(2) + 3x + 2)^(2))^(1/2) =
ln ( sqrt (x)) - ln (x^(2) + 3x + 2)

Answer 13

Alright, so you have \[\ln(\sqrt{x})-\ln(x^2+3x+2) \iff \ln(x^{1/2}) -\ln(x^2+3x+2)\]Which is correct \(\checkmark\)

Answer 14

So using the properties of logs,
ln (x+2) + ln (sqrt(x)) - ln (x^(2) + 3x + 2) would be
ln ( ((x+2)(sqrt(x))/((x)^(2)+3x+2)) ) ?

Answer 15

\[\ln(\frac{(x+2)\sqrt{x}}{ x^2 + 3x + 2 })\]

Answer 16

That's correct :)

Answer 17

\[\log(a) + \log(b) - \log(c) \iff \log\left(\frac{ab}{c}\right)\]

Answer 18

But now you need to simplify \(x^2+3x+2\)

Answer 19

\[\ln(\frac{ \sqrt{x}(x+2) }{ (x+1)(x+2) }) = \ln(\frac{ \sqrt{x} }{ x+1 })\]

Answer 20

Ohh haha my bad. Sorry.

Answer 21

Oh. Our teacher prefers using ln.

Answer 22

yeah, I just misread my own post, lol.

Answer 23

Thanks for explaining everything. :)

Answer 24

But that's correct. \[\ln\left(\frac{x^{1/2}}{x+1}\right) \iff \ln(x^{1/2}) -\ln(x+1) \iff \frac{1}{2}\ln(x) - \ln(x+1)\]

Answer 25

No problem :)