Question

I need to define this explicitly: ln(3-x) - 1/2(ln(y^2+16)) = C Some one give me a hand just to make sure I did my algebra correctly

Answer 1

for x or y ?

Answer 2

For y

Answer 3

\[\frac{ln(3-x) - 1}{2(ln(y^2+16))} = C\]
this?

Answer 4

Sorry lemme redo the equation. Lol it's a lot simpler than that

Answer 5

\[\ln (3-x) - 1/2(\ln(y ^{2}+16)) = c \]

Answer 6

\[ln (3-x) - ln(y ^{2}+16)^{1/2} = C\]
\[ln (3-x) = C+ ln(y ^{2}+16)^{1/2}\]
\[e^{ln (3-x)} = e^{C+ ln(y ^{2}+16)^{1/2}}\]
\[3-x = e^C e^{ln(y ^{2}+16)^{1/2}}\]
\[3-x = e^C (y^{2}+16)^{1/2}\]
so far?

Answer 7

Yeah looks good

Answer 8

But wait, how did you go from C + ln(whatever) to multiplying them together?

Answer 9

\[\left(3-x = e^C (y^{2}+16)^{1/2}\right)^2\]
\[(3-x)^2 = e^{2C} (y^{2}+16)\]
\[\frac{(3-x)^2}{e^{2C}} = y^2+16\]
\[\frac{(3-x)^2}{e^{2C}}-16 = y^2\]
\[\sqrt{\frac{(3-x)^2}{e^{2C}}-16} = y\]

Answer 10

shouldn't it be e^C +(y^2+16)^1/2

Answer 11

\[B^{a+b} = B^{a} B^{b}\]

Answer 12

B=e
a = C
b = ln(...)

Answer 13

so e^C + e^ln(y^2+16)^(1/2) = e^(C+(y^2+16)^(1/2)) ? That just doesn't quite seem right to me

Answer 14

\[\sqrt{\frac{(3-x)^2}{e^{2C}}-16} = y\]
\[\sqrt{\left(\frac{(3-x)}{e^{C}}\right)^2-(4)^2} = y\]
can we go further?

Answer 15

\[\huge e^{ln (3-x)} = e^{(C+ ln(y ^{2}+16)^{1/2})}\]
\[\huge e^{ln (3-x)} = e^{C}\ e^{ln(y ^{2}+16)^{1/2}}\]

Answer 16

Well C = ln(18) if what I did earlier in the problem was correct

Answer 17

that does help :)

Answer 18

when we "e" both sides; the RHS turns into a sum of exponents

Answer 19

I have kind of a side question. does 1/2(ln(y^2+16)) = ln(y+4) ?

Answer 20

im prone to say no; the sqrt of an sum of squares tends to be complex

Answer 21

sqrt doesnt distribute over addition; only multiplication

Answer 22

Well that's basically why I was getting wrong answers then because I put that in my problem haha

Answer 23

\[\sqrt{ab}=\sqrt{a}*\sqrt{b}\]
\[\sqrt{a+b}\ne\sqrt{a}+\sqrt{b}\]

Answer 24

Gotcha, I shoulda known that

Answer 25

I have a question, before you e'd both sides... wouldn't it have been easier to keep both logs on the same side? so you'd have like 3-x -1/2(y^2+16)?

Answer 26

perhaps, but i never really go for "easier" :) it would have amounted to the same thing in the end yes