If ln(y) = ln(x-1) + ln(x+1) + ln(c) then y = ?

Question

Massive problems with natural log (ln) here for school...how do I go about this??

Thanks!!

Wow thanks man i guess thats it...i just never knew that rule! How come you cant just put everything to the power of e, though? I.e. couldnt you just have made it e^(ln(y)) = e^(ln(x-1)) + e^(ln(x+1)) + e^(ln(c)) ??

Cool I get the first step then...but how do you justify that 'exponentiate both sides' move? If I may be so pestering :)

Answer 1

ln(y) = ln(x-1) + ln(x+1) + ln(c)

When logs are added, the arguments can be multiplied together under one log.

ln(y) = ln[(x-1)(x+1)c]

Exponentiate both sides.

y = (x-1)(x+1)c = c(x - 1)(x + 1)

Answer 2

ln(y) = ln(x-1) + ln(x+1) + ln(c)
ln(y) = ln[(x-1)(x+1)(c)]
y = (x-1)(x+1)(c)
y = c(x^2-1)
y = x^2 c - c

Answer 3

ln(y) = ln(x+1)+ln(x-1)+ln(c)
simplifying
ln(y) = ln(x+1)(x-1)(c)
ln(y) = ln(c*(x^2-1)
y=c*(x^2-1)
y=cx^2-c