# solve for x : lnx + ln(3x-2)=0

hala718 | Certified Educator

calendarEducator since 2008

starTop subjects are Math, Science, and Social Sciences

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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justaguide | Certified Educator

calendarEducator since 2010

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.

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