Question

write as a single log

1/2Ln(x^2+1)-(4Ln1/2)-(1/3Ln(x-4)

Answer 1

\[\frac{1}{2}\ln(x^2+1)-4\ln(\frac{1}{2})-\frac{1}{3}\ln(x-4) =\]
\[\ln((x^2+1)^{\frac{1}{2}})-\ln((\frac{1}{2})^{4})-\ln((x-4)^{\frac{1}{3}}) =\]

\[\ln\frac{(x^2+1)^{\frac{1}{2}}}{(\frac{1}{2})^{4}(x-4)^{\frac{1}{3}}}\]
wow that took a lot longer to type out than i thought >.>

Answer 2

You just need to know how to play with natural logs:
\[aln(b) = \ln(a^{b}), \ln(a)+\ln(b) = \ln(a*b), \ln(a)-\ln(b) = \ln(\frac{a}{b})\]

Answer 3

why is the 1/2^4 multiply with (x-4)^1/3 in the denominator

Answer 4

its all subtractions isnt it?

Answer 5

that is a little tricky, but i works out like this:
\[\ln(a)-\ln(b)-\ln(c) = \ln(a)-(\ln(b)+\ln(c)) = \ln(a)-\ln(b*c) = \ln(\frac{a}{b*c}) \]

Answer 6

i understand but dont really get it why is the negative turn into positive?

Answer 7

if i had something crazy like:
\[\ln(5)+\ln(10)-\ln(2)-\ln(6)+\ln(cat)-\ln(dog)\]
in the end it would just be:
\[\ln\frac{(5)(10)(cat)}{(2)(6)(dog)}\]
Group all the positive terms and thats the numerator, then group the negative terms and thats the denominator. For the negative terms, its like you are factoring out the (-1) and adding them together, thats why you multiply them, but that (-1) is there on the outside, so you divide by that product

Answer 8

hey the answer is \[Ln [16\sqrt{x^2+1}]/\sqrt[3]{x-4}\]

Answer 9

where does 16 comes from?

Answer 10

ah, they moved the (1/2)^4 to the top of the fraction, because:
\[(\frac{1}{2})^{4} = \frac{1}{2^{4}} = \frac{1}{16}\]

Answer 11

but its in the denominator, so:
\[\ln\frac{(junk)}{\frac{1}{16}(otherjunk) }= \ln\frac{16(junk)}{(otherjunk)} \]

Answer 12

you would multiply the top and bottom of that fraction by 16, getting rid of the fraction in the denominator, and putting a 16 in the numerator

Answer 13

All-in-all this was a pretty complicated problem >.<

Answer 14

yup! i understand it now! you are good thank you sir!

Answer 15

how about if i have

Answer 16

\[Ln^2(X-4)\]

Answer 17

hmm....i dont think ive ever seen something like that....so basically its:
\[\ln(\ln(x-4))\]
is that correct?

Answer 18

im not sure either i thought you put the 2 down

= 2Ln(x-4)

Answer 19

Oh, its this then:
\[\ln((x-4)^{2})\]
Thats when you can pull the 2 down and make it:
\[2\ln(x-4)\]

Answer 20

as you can see, understanding notation and interpretation is 99% of the battle lol

Answer 21

hehe yea so LN^2 is same as ln((x−4)^2)

Answer 22

i guess so o.O althought ive never seen:
\[\ln^{2}(whatever)\]
before, to each their own i suppose lol

Answer 23

how about e^(-Ln3) is this = to -3?

Answer 24

and e^((ln2)x) =??

Answer 25

close! so close, you have the right idea, you want to cancel out the ln with e. but lets look at this term a little closer:
\[-\ln(3)\]
If we use that rule:
\[aln(b)= \ln(b^{a})\]
then we have:
\[(-1)\ln(3) = \ln(3^{-1}) = \ln(\frac{1}{3})\]

Answer 26

So :
\[e^{-\ln(3)} = e^{\ln(\frac{1}{3})} =\frac{1}{3} \]

Answer 27

hmm...for the second...lets see:
\[e^{\ln(2)x} = (e^{\ln(2)})^{x} = (2)^{x} = 2^{x}\]
Another way to see it is:
\[e^{\ln(2)x} = e^{\ln(2^{x})} = 2^{x}\]

Answer 28

is there a rule to raise the 2 to power x?

Answer 29

Not really, its just an expression, or a function. not really much you can do with it, although it look a lot nicer than what we started with.

Answer 30

hehe yea thanks a lot!

Answer 31

no prob :)

Answer 32

have a good day

Answer 33

you too, good luck with your work :)

Answer 34

thanks