Question

Check my logarithm answers?

Answer 1

Combine into a single logarithm.
3log(x+y) + 2log(x-y) - log(x^2+y^2)
My answer: log (x+y)^3(x-y)^2 over (x^2+y^2)

Use rules of logarithms to expand.
log (x+3)^2 over (x-2)(x^2+5)^4
My answer: 2log(x+3) - log(x-2) + 4log(x^2+5)

I can make the equations if these are too hard to read as is...

Answer 2

note: \(\log(\dfrac{a}{bc})=\log(a)-[\log(bc)]=\log(a)-[\log(b)+\log(c)]\\=\log(a)-\log(b)-\log(c)\)

Answer 3

You forgot to distribute the negative.

Answer 4

On the first or the second problem? Thanks for answering!

Answer 5

Second

Answer 6

Gotcha! So the last term should be negative?
Like this? 2log(x+3) - log(x-2) - 4log(x^2+5)

Answer 7

\(\log(\dfrac{(x+3)^2}{(x-2)(x^2+5)^4})=\log((x+3)^2)-[\log((x-2)(x^2+5)^4)]\\=\log((x+3)^2)-[\log((x-2))+\log((x^2+5)^4)]\\=\log((x+3)^2)-\log((x-2))-\log((x^2+5)^4)\)

Answer 8

yes

Answer 9

Is that all I had wrong? Thanks

Answer 10

Very common mistake and I would look for something like that on the test :) I always put one on my tests.

Answer 11

Sneaky!

Answer 12

the rest looks good

Answer 13

Cool. I have two more, if you don't mind checking them as well.

Answer 14

I tell them in class, over and over, that I will put one on the test ....

Answer 15

close this and open a new one, I have to go soon. But many people here can help on this.

Answer 16

My guess is that you have it right and you can always check wolfram

Answer 17

Rewrite as an exponential equation.
ln(x+y) = 5
My answer: e^5 = x+y

Rewrite as a logarithmic equation.
s^4 = (a+-b)
My answer: log_2(a-b) = 4

Answer 18

Oh okay

Answer 19

Thanks for your help!

Answer 20

wait

Answer 21

should that s be a 2?

Answer 22

if so those are both correct @Abbles

Answer 23

Yes, it should be a 2. And thank you so much!