Sapling #18
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Re: Sapling #18
You raise it to e! just like to get rid of -log you raised it to 10, you raise the ln() to e. Hope this helps!
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Re: Sapling #18
First isolate ln(K) to one side. Then, to get K, you should do e^(whatever is on the other side).
Here's an example of how to do it with some random numbers:
ln(x) = 2.3/4
x = e^(2.3/4)
Here's an example of how to do it with some random numbers:
ln(x) = 2.3/4
x = e^(2.3/4)
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Re: Sapling #18
Hi! Ln is just log base e, so to get rid of it, just place whatever's on the opposite side of the equivalence sign as the power of e. Hope this helps!
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Re: Sapling #18
It was strange to me when I did it and got a very large number, but trust your gut for the math and you should be good! :)
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Re: Sapling #18
If you want to solve for what's inside the ln you need to take the reverse function of it: like diving is the reverse of multiplying, and - is to +.
The reverse of ln is e so make both sides of the equation e^ln(X)=e^#you get. This should help you solve it!
The reverse of ln is e so make both sides of the equation e^ln(X)=e^#you get. This should help you solve it!
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Re: Sapling #18
Hello,
For this one, you raise e to the value given for lnK. For example if lnK = 200 then K = e^200.
Hope this helps!
For this one, you raise e to the value given for lnK. For example if lnK = 200 then K = e^200.
Hope this helps!
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Re: Sapling #18
I would take the constant inside of the ln (like 40 if it was ln40) and then raise e to that power!
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Re: Sapling #18
First, Isolate the ln term and then take e to the power of that that ln(x) equals to get x
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