is ln(x) + ln(y) = ln(y/x)? and is e^(neg. infinity) = 0?

Question

pls help... i'm getting confused already
is e raised to negative infinity equal to 0?

Answer 1

Provided both x and y are positive,
ln x + ln y = ln (xy) according to the property of logarithm.

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e^(neg. infinity) is 0, taken as a limit.

Answer 2

no ln(x) + ln(y) = ln(xy)
it is -ln(x) + ln(y) = ln(y/x)

it is bad to write e^(-infinity) = 0, although it is true in a sense.
Properly you should say the limit as x tends to minus infinity of e^x is 0. Or:
lim_{x->-inf} e^x = 0

Answer 3

Question 1. NO. ln(x)+ln(y) = ln (x*y)
Question 2. NO. The value approaches zero.

Answer 4

because of the fact it leads to paradoxes. assume a million/0=infinity. Now bear in strategies that a million/2*2=a million,a million/.a million*.a million=a million,and a million/x*x=a million. a million/0*0=a million bear in strategies that any element situations 0 equals 0. Infinity*0=0 a million/0*0=infinity*0 a million=0

Answer 5

ya, what the guy say above if to get ln(y/x) its lny-lnx
as got the e^ neg infinity, well... pretty much, but if your doing caluclus, you ur not suppose to just say its infinity, its improper, you have to put lim as c goes to neg infinity..etcetc,

Answer 6

ln(x)+ln(y)=ln(xy)

For your second question, you must think of it as a limit, e^-infinity only approaches zero.

Answer 7

ln(x) + ln(y) = ln(x.y)
ln(x) - ln(y) = ln(x/y)

a number ^neg. infinity = 0 and "e" is a number
e=2.718

Answer 8

ln(x)+ln(y)=ln(x*y)
ln(x)-ln(y)=ln(x/y)

And yes, lim(exp(x), x --> -infinity)=0