solve for x : lnx + ln(3x-2)=0
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ln x + ln (3x-2) = 0
We will use the logarithm properties to solve for x.
We know that: log a + log b = log a*b
Then we will write:
ln x + ln (3x-2) = ln x*(3x-2) = 0
==> ln (3x^2-2x) = 0
Now we will rewrite into the exponent form.
==> 3x^2 - 2x = e^0 = 1
==> 3x^2 - 2x -1 = 0
Now we will use the formula to find the roots.
==> x1= (2 + sqrt(4+4*3) / 2*3 = (2+4)/6 = 1
==> x2= (2-4)/6 = -2/6 = -1/3 ( not defined for ln x)
Then the only solution is x= 1
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calendarEducator since 2010
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starTop subjects are Math, Science, and Business
We have to solve for x: ln x + ln(3x-2) = 0
ln x + ln(3x-2) = 0
we use the relation log a + log b = log a*b
=> ln x*(3x -2 ) = 0
=> ln 3x^2 - 2x = 0
we know that ln 1= 0
=> ln 3x^2 - 2x = ln 1
taking antilog of both the sides
=> 3x^2 - 2x = 1
=> 3x^2 - 2x - 1 = 0
=> 3x^2 - 3x + x - 1 = 0
=> 3x( x - 1) + 1( x - 1) = 0
=> ( 3x + 1)( x - 1) = 0
=> x = -1/3 and x = 1
But ln (-1/3) is not defined.
Therefore the only solution here is x = 1.