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Solve the logarithmic equation: ln(3x-4)=ln4-ln(x-1)

2007-04-18 17:24:01 · 4 answers · asked by xtaticlyme 2 in Science & Mathematics Mathematics

4 answers

ln (3x-4) = ln 4 - ln(x-1)
<=> ln (x-1) + ln (3x-4) = ln 4
<=> ln [(x-1)(3x-4)] = ln 4
<=> (x-1) (3x-4) = 4 and x-1 > 0, 3x-4 > 0
<=> 3x^2 -7x = 0 and x > 4/3
<=> x (3x - 7) = 0 and x > 4/3
<=> (x = 0 or 7/3) and x > 4/3
<=> x = 7/3.

2007-04-18 17:30:15 · answer #1 · answered by Scarlet Manuka 7 · 1 0

Use the houses of logarithms: lnA + lnB = ln(A*B), for any A and B more beneficial than 0 hence: lnx + ln(2x+a million) = ln[x*(2x+a million)] = ln[2x² + x] = 0 on condition that 0 = ln a million, we've: ln(2x² + x) = ln a million 2x² + x = a million 2x² + x - a million = 0 x = {-a million +_ sqrt [a million² - 4*2*(-a million) ] }/(2*2) = {-a million +_ sqrt(a million + 8)}/4 = {-a million +_ sqrt 9}/4 = {-a million +_ 3}/4 x = 2/4 = 0.5 OR x = -4/4 = -a million even if, the condition of existence of logarithms is that each and each one numbers "interior" the logarithm must be effectual. So x = -a million isn't a conceivable answer, because lnx = ln (-a million) does no longer EXIST. answer: x = 0.5 = a million/2

2016-12-04 07:14:37 · answer #2 · answered by janta 4 · 0 0

ln(3x-4)=ln4-ln(x-1) use ln a -ln b=ln (a/b)
ln (3x-4)=ln (4/(x-1)) take exponential of each side
3x-4=4/(x-1) multiply both sides by x-1
(3x-4)(x-1)=4 use FOIL
3x^2-3x-4x+4=4 subtract 4 from each side
3x^2-7x=0
x(3x-7)=0
x=0
3x-7=0
x=7/3
0, 7/3

2007-04-18 17:39:06 · answer #3 · answered by yupchagee 7 · 1 0

ln (3x-4) = ln 4 - ln(x-1)
=> ln (x-1) + ln (3x-4) = ln 4
=> ln [(x-1)(3x-4)] = ln 4
=> (x-1) (3x-4) = 4
=> 3X^2 -7X+4=4
=> X(3X-7)=0
=> X=0 or 3X-7=0
=> X=0 or X=7/3
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2007-04-18 17:52:27 · answer #4 · answered by vishnu4100 3 · 0 0

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