Question

Write the expression below as a single logarithm:
(1/3)Inx + 4In(yz) - 3In(x^2yz).

How do I begin? Write the expression below as a single logarithm:
(1/3)Inx + 4In(yz) - 3In(x^2yz).

How do I begin? @Mathematics

Answer 1

you move the numbers in front to become powers of the expressions you're taking ln of. i.e. \[blna=\ln(a^b)\]

Answer 2

(1/3)^x + 4^yz - 3^x^2yz

Is that what you mean?

Answer 3

yes is right but you know again that lna +lnb=ln(a*b)

Answer 4

no, \[(1/3) \ln x = \ln [x^(1/3)]\] and so on. its a logarithm law :)

Answer 5

and lna - lnb=ln(a/b)

Answer 6

oops, equation messed up lol. correction: \[{1/3}\ln x =\ln [x^{(1/3)}]\]

Answer 7

you do that for all three logs, and then use the laws that jhonyy9 mentioned to rewrite the expression as a single logarithm to the base e :)

Answer 8

\[ In[x^(1/3)] + In[(yz)^4] - In[(x^2yz)^3]\]

Answer 9

yup thats right :)

Answer 10

ln((cuberootx)(yz)4)/(x2yz)3) = ln(cuberoot(x)(yz)/(x2))

Answer 11

now use the fact that ln x +ln y = ln xy and ln x - ln y = ln (x/y)

Answer 12

\[In[x^(1/3)] + In[(yz)^4] = In[(x^2yx)^3]\]

There's the first one.

Answer 13

unfortunately, you can't set the one side equal the other, because the whole expressions isn't equal to zero.

Answer 14

I have never struggled so much in a math class before.

Answer 15

what class is it?

Answer 16

It's called College Algebra, but is actually Pre-Calculus.

Answer 17

oh that class was tough ><

Answer 18

Tell me about it.

Answer 19

it gets better though, once you get the hang of everything :)

Answer 20

I thought that I was going to happen, but now I'm in my final week

Answer 21

ouch. well on the bright side, you won't have to deal with this anymore after the week is through :D

Answer 22

lol. Yeah, with the possibility of having to take this over again! I'm really looking forward to it. :D

Answer 23

i wish you luck with that possibility :)