is it correct if i write it like this to find x:
l0.5 = l_0e^(-0.035*x)?
#logarithms
formal dust
fleet heathBOT
is it correct if i write it like this to find x:
l*0.5 = l_0*e^(-0.035*x)?
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meager cove
is it correct if i write it like this to find x:
l*0.5 = l_0*e^(-0.035*x)?
Express x first, then substitute the constant.
meager cove
yes but do i find x like this?
No, not quite. We have:
I = I0 e^(-kx)
We want the distance x at which the intensity halves. So, I = (1/2)I0. So:
(1/2)I0 = I0 e^(-kx)
And now you can solve for x.
Or you can solve for x first, then substitute I/I0 = 1/2. Doesn't matter.
formal dust
No, not quite. We have:
I = I0 e^(-kx)
We want the distance x at which the inten...
aaa so instead of l i have to write l_0?
or another constant
like “a” for instance
meager cove
aaa so instead of l i have to write l_0?
Well, we want to find the distance at which the intensity halves, so I = (1/2)I0.
formal dust
Well, we want to find the distance at which the intensity halves, so I = (1/2)I0...
so like this?
meager cove
so like this?
Yes. More generally, for exponential decay the distance at which the intensity halves would be x = ln(2)/k.
formal dust
Yes. More generally, for exponential decay the distance at which the intensity h...
huh
why divided by k
meager cove
why divided by k
Well, again, try solving I = I0 e^(-kx) for x.
formal dust
okok
meager cove
A common mistake I see people doing is substituting numbers first, then solving. That's now how you solve problems in physics or chemistry.
I wonder why, though...
formal dust
Well, again, try solving I = I0 e^(-kx) for x.
but how do i get x alone
on one side
then
meager cove
Well, how do you solve exponential equations?
formal dust
Well, how do you solve exponential equations?
using logarithm or ln
meager cove
using logarithm or ln
Not sure what you mean. ln is natural logarithm (base e).
meager cove
but this answer would be correct?
Well, if you didn't recognize that this is ln(2)/k, than that isn't very useful information 😅
formal dust
so i cam use log base e
formal dust
then i can use ln
meager cove
and on the other side i get e^0.035*x
e^(kx).
formal dust
The formula contains I, not l.
ohh ok
formal dust
e^(kx).
well yes
and then i take ln on both sides
and i get kx
on one side
or no
formal dust
And on the other?
right?
meager cove
Same thing.
formal dust
yess
formal dust
Then, what will x be?
right?
formal dust
yes
formal dust
ok
meager cove
We wrote the initial formula a bit wrong. Should be I = I0 e^(-kx).
Then x = -(1/k)ln(I/I0).
formal dust
We wrote the initial formula a bit wrong. Should be I = I0 e^(-kx).
Then x = -(1...
so divided by negative k
is the only difference
meager cove
yes the k is negative
Well, k itself is positive, but there's a minus sign in front.
meager cove
so divided by negative k
Better to write the minus in front, not in the denominator.
formal dust
okok
meager cove
Anyway, if x = -(1/k)ln(I/I0), what do we get when I/I0 = 1/2?
formal dust
well i got x=-100/7
meager cove
I mean as a formula.
meager cove
Where did the logarithm go?
meager cove
Well, yes.
But let's not use decimal fractions.
We get x = -ln(1/2)/k.
How can we simplify the numerator?
nimble blade
@formal dust how many logs are they making u do hahaha
meager cove
Well, they are quite important.
formal dust
+close