Solve for x in lnx+ln(x-6)=1?
回答 (4)
2009-02-07 3:35 pm
✔ 最佳答案
lnx+ln(x-6)=1since ln A + ln B = ln(A*B)
ln (x(x-6)) = 1
ln e=1
so
ln (x(x-6)) = 1
ln e
x(x-6) = e
x^2 -6x -e=0
If you need to solve for x we will need to approximate e and use the quadratic formula, or my choice, simply solve graphically.
x ≈ -.4231976, 6.4231976
參考: TI - 83 plus
2009-02-08 12:07 am
ln(x) + ln(x - 6) = 1
ln[x(x - 6)] = 1
x(x - 6) = e^1
x^2 - 6x - e = 0
x = [-b 屉(b^2 - 4ac)]/2a
a = 1
b = -6
c = -e
x = [6 屉(36 + 4e)]/2
x = {6 屉[4(9 + e)]}/2
x = [6 ±2â(9 + e)]/2
x = 3 屉(9 + e)
â´ x = 3 ±â(9 + e)
ln[x(x - 6)] = 1
x(x - 6) = e^1
x^2 - 6x - e = 0
x = [-b 屉(b^2 - 4ac)]/2a
a = 1
b = -6
c = -e
x = [6 屉(36 + 4e)]/2
x = {6 屉[4(9 + e)]}/2
x = [6 ±2â(9 + e)]/2
x = 3 屉(9 + e)
â´ x = 3 ±â(9 + e)
2009-02-07 11:45 pm
lnx+ln(x-6)=1
(10^x)(10^(x-6)=10^1
x(x-6)=1
x=0 x-6=0
x=6 answer//
(10^x)(10^(x-6)=10^1
x(x-6)=1
x=0 x-6=0
x=6 answer//
2009-02-07 11:42 pm
using property lna+lnb=ln(ab) we have.....
ln(x*(x-6))=1
x^2-6x=e^1
x^2-6x-e=0
apply sri dharacharya method to solve quadratic eq.......
x = { -(-6) + sqrt[ (-6)^2 - 4(1)(e) ] } /2(1) or
{ -(-6) - sqrt[ (-6)^2 - 4(1)(e) ] } /2(1)
take e=2.718 to calculate the final answer!!!!!
ln(x*(x-6))=1
x^2-6x=e^1
x^2-6x-e=0
apply sri dharacharya method to solve quadratic eq.......
x = { -(-6) + sqrt[ (-6)^2 - 4(1)(e) ] } /2(1) or
{ -(-6) - sqrt[ (-6)^2 - 4(1)(e) ] } /2(1)
take e=2.718 to calculate the final answer!!!!!
收錄日期: 2021-05-01 11:57:47
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20090207072705AAMWVeP