Why is ln^2(x)=2lnx?

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answers:
say: Troll + Lazy @$$ who won't do his own homework = Dork Boy.
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Vaman say: This is not correct. It should be
ln x^2. ln^2 x has no meaning.
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1234567 say: It's not.
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Raymond say: The basic rule of exponents (and, therefore, of logs) is
when you multiply powers of a base, you add the power.

Since a log (by definition) is the power itself in a given base (the base of the log), the log of a multiplication is the sum of the logs.

log(ab) = log(a) + log(b)

log(x^2) = log(xx) = log(x) + log(x) = 2log(x)
(in ANY base, as long as you keep the same base throughout the calculation).

However, squaring the log itself is different.

ln^2(x) is not the same as ln(x^2)

Let's say x=7
ln(49) = ln(7^2) = 2ln(7)
we check
3.891820... = 2 * 1.945910...
yep! it works.
However,
[ln(7)]^2 = 3.7865663... = ln(44.104698...)
not equal to ln(49)
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DWRead say: The exponent is in the wrong place.

ln(x²) = 2ln(x)
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Ray S say: It's not.

ln²(x) = ln(x) * ln(x)
But,
ln(x²) = 2 ln(x)
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Alvin say: It isn't
it is only true when x =1 and x = 2e
ln^2(1) = 2*ln(1) = 0

ln^2(2e) =2*ln(2e) = 4
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billrussell42 say: it's not
but
ln(x²) = 2 ln(x)

ln²(x) = ln(x)ln(x) which cannot be reduced.
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ted s say: it is NOT the same....you have [ ln x ]² ╪ 2 ln x = ln ( x² )

depending upon how the ln function was introduced the answers are different...if you used : y = ln x if and only if x = e^y then x² = e^(2y) == > iff ln x² = 2y = 2 ln x
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say: it isn't, ln(x^2) = 2ln(x) and you can use exponent laws to deduce that
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Ocate say: ln ² (x) = 2 lnx
it is similar to log ² x = 2 log x
it is similar to log ² (100) = 2 (2) and 2 log (100) = 2 (2) = 4
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