1) ln()=2.
We know: ln()=n ln(x)
-> 2ln(x)=ln(2) --> ln(x)=. --> x=
.
Is that correct? Because my booklet, again, says something different ;)
2) are generally all logarithmic rules appliable for ln?
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1) ln(
We know: ln(
-> 2ln(x)=ln(2) --> ln(x)=
Is that correct? Because my booklet, again, says something different ;)
2) are generally all logarithmic rules appliable for ln?
I obtained a this:
Multiple by 2
Raise to the e
1. answered above.Quote:
Originally Posted by Schdero
1) ln(
We know: ln(
-> 2ln(x)=ln(2) --> ln(x)=
Is that correct? Because my booklet, again, says something different ;)
2) are generally all logarithmic rules appliable for ln?
2. yes because ln(x) is the same thing as
Yes sure, sry i meant ln(x^2)=2, not x^1/2. But thats no big change, because in my booklet the solution is +/-e and this is nonsense, isn't it?
You can't have a negative value in the natural function since the domain is (0,infinity)
-e=-2.78......