how does 3ln(x²) - 3ln|x²-4y²| = 8ln|x| + C turn into (x²-4y²)³x² = C?

Question

how does 3ln(x²) - 3ln|x²-4y²| = 8ln|x| + C turn into (x²-4y²)³x² = C

Not sure how to do this.

please show step by step

Answer 1

3ln(x^2) = ln(x^6)

3ln|x^2-4y^2| = ln((x^2-4y^2)^3)

8ln|x| = ln(x^8)



now we have



ln(x^6) - ln((x^2-4y^2)^3) - ln(x^8) = C



which simplifies to



ln (1/(x^2*(x^2-4y^2)^3)) = C

1/(x^2*(x^2-4y^2)^3) = e^C



but e^C is still a constant, so we'll just call it C still



1/(x^2*(x^2-4y^2)^3) = C



take reciprocal of both sides



x^2*(x^2-4y^2)^3 = 1/C



again, 1/C is a constant, so it's still just C



x^2*(x^2-4y^2)^3 = C



the C just means it's a constant, the C in the first equation isn't equal to the C in the second, if you want them to be the same then you would use e^(-C) in the second, but it's generally accepted that C is a constant.

Answer 2

3ln(x²) - 3ln|x²-4y²| = 8ln|x| + C



3ln (x^2) = ln (x^2)^3 = ln (x^6)

3 ln |x^2 - 4y^2| = ln |x^2 - 4y^2|^3

8 ln | x | = ln x^8



ln A - ln B = ln (A / B)



so 3ln(x²) - 3ln|x²-4y²| = 8ln|x| + C = ln [x^6 / ((x^2 - 4y^2)^3 x^8)) = ln [1 / [x^2 (x^2 - 4y^2)^3]

= - ln x^2 (x^2 - 4y^2)^3]



so -ln [(x^2 - 4y^2)^3 x^2) = C ==> ln [(x^2 - 4y^2)^3 x^2) = -C

e^(-C) = (x^2 - 4y^2)^3 x^2



but e^(-C) is another constant, so you now have

(x^2 - 4y^2)^3 x^2 = D



(this will not be the same as the original constant)

D = e^(-C)

Answer 3

3ln(x²) - 3ln|x²-4y²| = 8ln|x| + C

3ln(x²) - 3ln|x²-4y²| - 8ln|x| = C



Remember the property of logarithms:

nln a = ln a^n, so:



3ln(x²) - 3ln|x²-4y²| - 8ln|x| = C

ln((x²)³) - ln(|x² - 4y²|³) - ln(|x|^8) = C



Also, remember that x² = |x|², and |x|^8 = (|x|^2)^4 = x^8, so:



ln(x^6) - ln(|x² - 4y²|³) - ln(x^8) = C



Lastly, remember that:

ln a + ln b = ln(ab), and ln a - ln b = ln(a/b), so:



ln(x^6) - ln(|x² - 4y²|³) - ln(x^8) = C

ln(x^6/x^8) - ln(|x² - 4y²|³) = C

ln(1/x²) - ln(|x² - 4y²|³) = C

ln((1/x²)/|x² - 4y²|³) = C

ln(1/|x² - 4y²|³x²) = C

e^(ln(1/|x² - 4y²|³x²)) = e^C, if C is a constant, then e^C is also a constant:

1/|x² - 4y²|³x² = C

|x² - 4y²|³x² = 1/C, if C is a constant, then 1/C is also a constant:

|x² - 4y²|³x² = C



Notice that while C ≠ e^C ≠ 1/(e^C), they are all constant. C is supposed to represent an arbitrary constant number, so on that case we write C = e^C = 1/(e^C), since they all represent arbitrary constants, we make them assume any number we want.

Answer 4

3ln(x^2) - 3ln(x^2 - 4y^) = 8ln(x) + C

ln(x^2)^3 - ln((x^2 - 4y^2)^3) = ln(x^8) + c



ln[x^6] - ln((x^2 - 4y^2)^3 - ln(x^8) = C



ln[x^6/[(x^2 - 4y^2)^3*x^8)] = C



ln[1/(x^2 + 4y^2)^3 * x^2] = C



-ln[x^2 (x^2 - 4y^2)^3] = -C but this is a constant so replace with a generic C



ln(x^2 (x^2 - 4y^2)^3) = C



rewrite in exponential form:



x^2(x^2 - 4y^2)^3 = e^C which is still a constant



x^2(x^2 - 4y^2)^3 = C

Answer 5

remember that for any base

log a + log b = log (ab)

log a - log b = log (a/b)

n log a = log aⁿ



in particular

- log a = log a⁻¹ = log (1/a)



and

e^ln(x) = x

ln(e^x) = x



3ln(x²) - 3ln|x²-4y²| = 8ln|x| + C



ln(x²)³ - ln|x²-4y²|³ - ln x⁸ = C



ln x⁶ - ln|x²-4y²|³ - ln x⁸ = C



ln x⁶ /(|x²-4y²|³ x⁸) = C



x⁶ and x⁸ simplify



ln 1 /(|x²-4y²|³ x²) = C



multiply both sides by (-1)



- ln 1 /(|x²-4y²|³ x²) = -C



ln [1 /(|x²-4y²|³ x²)]⁻¹ = -C



ln (|x²-4y²|³ x²) = - C



|x²-4y²|³ x² = e^(-C)



|x²-4y²|³ x² = K



in the last line using the same letter C can be confusing