# Solve for x ? 2lg x=lg(3x-2)

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2lg x = lg(3x-2)

lg (x)^2 = lg (3x -2)

Then:

x^2 = 3x -2

==> x^2 -3x +2 =0

Factorize:

==> (x-2)(x-1)

==> x1= 2

and x2= 1

To check:

2lg (2) = log(3(2)-2)

log(4) = log (4)

For x2= 1:

2log(1) = log (3(1)-2)

2 (0) = log (1)

0=0

2lgx =lg(3x-2). To solve for x.

Solution:

nlgx = lgx^n. So 2lgx = lgx^2. So the given equation becomes:

lgx^2 =lg(3x-2). Take antilog.

x^2 = 3x-2.

x^2-3x+2 = 0.

x-2x -x +2 = 0

x(x-2)-1(x-2) = 0

(x-2)(x-1) = 0

x-2 = 0, x-1 = 0

x-2, x=1

First, we'll use the power property of logarithms for the left term:

2lg x = lg(x^2)

We'll re-write the equation:

lg(x^2) = lg(3x-2)

We'll use the one to one property:

x^2 = 3x-2

We'll move all termsto the left side:

x^2 - 3x + 2 = 0

We'll use the quadratic formula:

x1 = [3+sqrt(9-8)]/2

x1 = (3+1)/2

x1 = 2

x2 = 1

We'll verify the results into equation:

2lg x1=lg(3x1 - 2)

2lg2 = lg(6-2)

lg 2^2 = lg4

lg4 = lg4

2lg x2=lg(3x2 - 2)

2lg 1=lg(3-2)

2*0 = lg 1

0 = 0

So, both values of x are the solutions of the equation:

x1 = 2 and x2 = 1