log + ln + e

2011-03-12 8:52 am
1. log base 5 (2x+1)= 1 + log base 5 x
2. log base 3 (2x+1)=2- log base 3 (x-1)
3. log base 2 (x+3)+ log base 2 (x-1) =5
4. log base 2 (2x+1) +log bsse 2 (x-3) =2
5. ln(x+1)-lnx = ln3
6.ln((5x+1) - ln(x-3)=ln2
7.ln x + ln(x-1)= ln 6
8.ln(6x-2)+1 = ln(2x+7)
9. ln x -ln(x-4) =3
10. 2 ln(x+3) - ln(3x+4) = ln 4

consider the exponential equation as ' quadratic equation' and solve by substitution

1. e^2x - 4e^x -5 =0
2.e^2x -8 e^x -9 =0
3.e^2x -e^x -12 =0
4.e^x - 5/(e^x) =4

回答 (1)

2011-03-12 11:29 am
✔ 最佳答案
1.
log5(2x + 1) = 1 + log5x
log5(2x + 1) = log55 + log5x
2x + 1 = 5x
3x = 1
x = 1/3


2.
log3(2x + 1) = 2 - log3(x - 1)
log3(2x + 1) + log3(x - 1) = 2log33
(2x + 1)(x - 1) = 3^3
2x^2 - x - 10 = 0
(2x - 5)(x + 2) = 0
x = 5/2 or x = -2 (rejected for log(-2 - 1) is undefined)


3.
log2(x + 3) + log2(x - 1) = 5
log2(x + 3) + log2(x - 1) = 5log22
(x + 3)(x - 1) = 2^5
x^2 + 2x - 35 = 0
(x - 5)(x + 7) = 0
x = 5 or x = -7 (rejected for log(-7 - 1) is undefined)


4.
log2(2x + 1) + log2(x - 3) = 2
log2(2x + 1) + log2(x - 3) = 2log22
(2x + 1)(x - 3) = 2^2
2x^2 - 5x - 7 = 0
(2x - 7)(x + 1) = 0
x = 7/2 or x = -1 (rejected for log2(-1 - 3) is undefined)


5.
ln(x + 1) - lnx = ln3
ln(x + 1) = ln3 + lnx
x + 1 = 3x
2x = 1
x = 1/2


6.
ln(5x + 1) - ln(x - 3) = ln2
ln(5x + 1) = ln2 + ln(x - 3)
5x + 1 = 2(x - 3)
5x + 1 = 2x - 6
3x = -7
x = -7/3 (rejected for log(-7/3 - 1) is undefined)
Hence, there is no solution.


7.
lnx + ln(x - 1) = ln6
x(x - 1) = 6
x^2 - x - 6 = 0
(x - 3)(x + 2) = 0
x = 3 or x = -2 (rejected for ln(-2) is undefined)


8.
ln(6x - 2) + 1 = ln(2x + 7)
ln(6x - 2) + lne = ln(2x + 7)
ln(6x - 2)e = ln(2x+7)
6ex - 2e = 2x + 7
(6e - 2)x = 2e + 7
x = (2e + 7)/(6e - 2)


9.
lnx - ln(x - 4) =3
lnx = 3lne + ln(x - 4)
lnx = lne^3(x - 4)
x = e^3(x - 4)
x = e^3x - 4e^3
(e^3 - 1)x =4e^3
x = 4e^3/(e^3 - 1)


10.
2ln(x + 3) - ln(3x + 4) = ln4
ln(x + 3)^2 = ln4 + ln(3x + 4)
(x + 3)^2 = 4(3x + 4)
x^2 + 6x + 9 = 12x + 16
x^2 - 6x - 7 = 0
(x - 7)(x + 1) = 0
x = 7 or x = -1


1.
Put u = e^x
e^2x - 4e^x - 5 =0
u^2 - 4u - 5 = 0
(u - 5)(u + 1) = 0
u = 5 or u = -1
e^x = 5 or e^x = -1 (rejected)
x = ln5


2.
Put u = e^x
e^2x - 8e^x - 9 = 0
u^2 - 8u - 9 = 0
(u - 9)(u + 1) = 0
u = 9 or u = -1
e^x = 9 or e^x = -1 (rejected)
x = ln9


3.
Put u = ex
e^2x - e^x - 12 = 0
u2 - u - 12 = 0
(u - 4)(u + 3) = 0
u = 4 or u = -3
e^x = 4 or e^x = -3 (rejected)
x = ln4


4.
Put u = e^x
e^x - 5/(e^x) = 4
[e^x - 5/(e^x)]e^x = 4e^x
(e^x)2 - 4e^x - 5 = 0
u^2 - 4u - 5 = 0
(u - 5)(u + 1) = 0
u = 5 or u = -1
e^x = 5 or e^x = -1 (rejected)
x = ln5
參考: miraco


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