Inactive Tutor answered • 09/21/16
Tutor
New to Wyzant
The distribution law for logarithms deals with
multiplication, not addition..
ln(ab) = ln(a)+ln(b)
But ln(a+b) ≠ ln(a) + ln(b)
Arturo proved this generically... It can also be shown simply
by putting in numbers for a, b
ln(2+3) = ln(5) ≈ 1.609438
ln(2)+ln(3) ≈ 1.791759
Thus ln(2+3) ≠ ln(2)+ln(3)
Inactive Tutor
Just be careful with the choice of numbers to sub! Example: 2 and 2
ln(2 + 2) = ln4
ln2 + ln2 = 2 ln2 = ln(22) = ln4
That is a rare case when the log of the sum = sum of the logs.
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09/21/16
Alex R.
Oh! I see a pattern here, you could argue the same thing with 2+2=4 vs 2*2=4. Are those are the only two subs that would ruin the rule?
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09/21/16
Inactive Tutor
To ruin the rule with some x and y, you need for x + y = xy to be true.
If so, you can show that x = y / (y - 1) for y > 1, since both x and y must be positive.
Example:
Pick y = 10. Then x = 10/9.
ln(x + y) = ln(10 + 10/9) ≅ 2.4079
ln(x) + ln(y) = ln10 + ln(10/9) ≅ 2.4079
It looks like as long as x + y = xy, ln(x + y) = lnx + lny, so there is an entire family of subs that for which ln(x + y) = lnx + lny.
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09/22/16
Alex R.
Fascinating! That makes sense since ln(2 + 2/1) ≅ 1.386 and ln(2) + ln(2/1) ≅ 1.386. It would seem that 2 and 2 are the only numbers that are the same AND fit into this family. It's crazy to know there is such a simple rule to find a whole group of decimals you can use. How did you figure this rule out (if it's not too complicated hah).
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09/22/16
Inactive Tutor
From proving by contradiction that ln(x + y) ≠ ln(x) + ln(y)
In the proof, we derived the necessary condition for ln(x + y) = ln(x) + ln(y), which is that x + y = xy, which is not true in general over the domain of ln(x).
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09/22/16
Alex R.
ln(1+1) ≈ .693
ln(1) + ln(1) = 0
09/21/16