is ln e^ (ab) = ab or a+b?
2006-09-02 10:51:59 · 9 answers · asked by th1a90 3 in Science & Mathematics ➔ Mathematics
ln e^ (ab) = ab*ln(e) by one of the laws of logarithms, and since ln(e) =1, then you can conclude that
ln e^ (ab) = ab
Also ln e^ (ab) means that you are looking for the exponent to which you have to raise the base, e, to get e^(ab), which obviously is ab.
2006-09-02 11:01:21 · answer #1 · answered by just♪wondering 7 · 0⤊ 0⤋
The natural log of a number is the inverse operation of raising e to the same number. You can think of them as canceling one another out.
ln e^(ab) = ab.
If you're not sure, assign values to a and b, and see what happens. Let a = 2 and b = 3. ab = 6 and a + b = 5. Find ln e^(6) on your calculator, and if the answer is 6, it's ab. If it turns out to be 5, it's a + b.
ln e^(ab) = ln e^6 = ln(403.4288) = 6. It's ab.
2006-09-02 13:47:16 · answer #2 · answered by Anonymous · 0⤊ 0⤋
It just equals ab
Since
ln(e^(ab))=ab*ln(e)
ln(e)=1
so
ab*ln(e)=ab
2006-09-02 11:52:54 · answer #3 · answered by Scott S 4 · 0⤊ 0⤋
ln e^ (ab) =(ab) ln e = (ab)(1) = ab
2006-09-02 10:55:12 · answer #4 · answered by fcas80 7 · 0⤊ 0⤋
ln(x) is the inverse of e^x, so ln(e^(ab)) = ab.
2006-09-02 11:22:41 · answer #5 · answered by john 3 · 0⤊ 0⤋
neither e^(ab) =e^a x e^(a(b-1))
hope this helped :)
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2006-09-02 10:56:08 · answer #6 · answered by absoluteao 3 · 0⤊ 0⤋
ln(e^(ab)) = ab
the ln cancels out the e^.
2006-09-02 10:58:15 · answer #7 · answered by smartee 4 · 0⤊ 0⤋
the second one
2006-09-02 10:53:17 · answer #8 · answered by Spaghetti MY 5 · 0⤊ 0⤋
ln_e(ab) = ln_e(a) + ln_e(b)
2006-09-02 12:46:21 · answer #9 · answered by Anonymous · 0⤊ 0⤋