2006-10-17 06:40:38 · 10 answers · asked by Cute B 1 in Science & Mathematics ➔ Mathematics
=ln(x) - ln(e) =ln(x) - 1
2006-10-17 06:49:35 · answer #1 · answered by ? 7 · 0⤊ 0⤋
INDICES RULE: whilst dividing x^a by employing x^b, subtract skill 'b' from 'a' to get the respond, x^c. as a consequence, d^5/d^3 may well be skill '5' minuspower '3' to get skill '2' or d^2. a million) that's in undemanding terms multiplying each and every ingredient contained interior the brackets by employing what's outdoors of it. 5 x 'x' = 5x. 5 x 4 = 20 4 x 'x' = 4x. 4 x -6 = -24. (undergo in innovations, a detrimental x constructive finally leads to a detrimental answer) So increasing question a million may well be 5x+20+4x-24. To simplify, do precisely the calculations. 5x+4x = 9x. 20-24 = -4. So the respond to question a million may well be 9x-4. For question 2, that's an comparable theory, yet they have thrown a skill in to attempt and throw you off. For the 1st bracket, that's 3x^2 multiplyed by employing x to get 3x^3. on the grounds which you're multiply x by employing x^2, you upload the powers jointly. 3x^2 x 4 = 12x^2. the 2d bracket, decrease back tries to throw you off. that's undemanding as quickly as you think of of roughly it. 3 x 'x^3' may well be 3x^3. 3 x -2 = -6 (decrease back, positives x negatives equivalent detrimental numbers) Your strengthen could be: 3x^3 + 12x^2 + 3x^3 - 6. it isn't any further available to function powers jointly. rapid occasion. 2x2x2 = 2^3 = 8. 2x2=2^2=4. 8+4=12. 12 isn't 2^4. the sole factor you will be able to do is 6x^3, on the grounds which you have 2 a great form of '3x^3'. Your answer for question 2, may well be 6x^3 + 12x^2 - 6. with any success I truly have been given all that precise. i've got have been given located some smart factors in for you.
2016-12-26 21:40:12 · answer #2 · answered by ? 3 · 0⤊ 0⤋
using the formula ln(u/v) = ln u - ln v
ln(x/e) = ln x - ln e = ln x -1 (ln e = 1)
2006-10-17 06:49:52 · answer #3 · answered by Mein Hoon Na 7 · 0⤊ 1⤋
ln(x)-1
2006-10-17 06:50:27 · answer #4 · answered by 1,1,2,3,3,4, 5,5,6,6,6, 8,8,8,10 6 · 0⤊ 1⤋
=ln(x) - ln(e) =ln(x) - 1
2006-10-17 08:08:10 · answer #5 · answered by Anonymous · 0⤊ 0⤋
ln(x/e) = ln(x) - ln(e) = ln(x) - 1
2006-10-17 07:35:57 · answer #6 · answered by Anonymous · 0⤊ 0⤋
ln(x) - ln(e) = ln(x) - 1.
2006-10-17 07:20:08 · answer #7 · answered by steiner1745 7 · 0⤊ 0⤋
ln(x/e) = ln(x) - ln(e)
now ln(e) = 1
so ln(x/e) = ln(x) - 1
2006-10-17 06:50:14 · answer #8 · answered by Stewart H 4 · 0⤊ 1⤋
What do you mean by that, I don't undersand your question?
2006-10-17 07:08:04 · answer #9 · answered by Anonymous · 0⤊ 0⤋
what is exactly do you want?
2006-10-17 06:49:47 · answer #10 · answered by openpsychy 6 · 0⤊ 2⤋