# Solve the equation 1/(1+lgx) + 3/(3+lgx)=2 .

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1/(1+ lgx) + 3/(3+lgx) =

First we will rewrite using the common denominator.

==> (3+lgx) + 3(1+lgx) = 2(1+lgx)(3+lgx)

Now we will expand the brackets.

==> 3+ lgx + 3 + 3lgx = 2(3 +lgx + 3lgx + (lg x)^2

==> 6 + 4lg x = 6 + 8lg x +2 (lgx)^2

==> 2(lg x)^2 + 4lgx = 0

==> 2lgx (lgx + 2) = 0

==> **lgx = 0 ==> x =1**

**==> lgx = -2 ==> x = 10^-2 = 1/100**

**==> x= { 1, 0.01}**

We'll impose the constraint of existence of logarithms:

x>0

The solution of the equation has to be in the interval (0 ; +infinite).

Now, we'll solve the equation. For this reason, we'll multiply both sides by the product (1+lgx)(3+lgx).

(3 + lgx) + 3(1 + lgx) = 2(1+lgx)(3+lgx)

We'll remove the brackets both sides:

3 + lgx + 3 + 3lgx = 6 + 2lgx + 6lgx + 2(lg x)^2

We'll combine like terms:

6 + 4lgx = 6 + 8lgx + 2(lg x)^2

We'll divide by 2 and we'll move all terms to the left side:

3 + 2lgx = 3 + 4lgx + (lg x)^2

4lgx + (lg x)^2 - 2lgx = 0

We'll combine like terms:

(lg x)^2 + 2lgx = 0

We'll factorize by lg x:

lg x(lg x + 2) = 0

lgx = 0

x = e^0

**x1 = 1**

lgx = -2

**x2** = 10^-2 = **1/10^2**.