Question 1171432
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#1. Log2(3x-7)+log2(x+2)=log2(x+1)

#2. Log2(3x+1)-log2(2-4x)>log2(5x-2)
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      The solution to equation  #1  in the post by  @CPhill,  giving the answer  x = 3, 


        is  TOTALLY,  GLOBALLY  and  FATALLY  incorrect.



<pre>
To check, it is enough to substitute x= 3 into equation #1.
You will get then in the left side

    log_2_(3*3-7) + log_2_(3+2) = log_2_(2) + log_2_(5) = log_2_(2*5) = log_2_(10);

in the right side 

    log_2_(3+1) = log_2_(4),


and even by unarmed eyes, you see that the left side is not equal to the right side.
</pre>

Contradiction which ruins the solution by @CPhill into dust.



&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;Below is my correct solution.



<pre>
Equation  log_2_(3x-7) + log_2_(x+2) = log_2_(x+1)  in its domain implies

    (3x-7)*(x+2) = x+1

    3x^2 - x - 14 = x+1

    3x^2 - 2x - 15 = 0.


The discriminant is  b^2 - 4ac = (-2)^2 - 4*3*(-15) = 4 + 180 = 184.

The discriminant is not a perfect square - so, the equation is not factorable.


Use the quadratic formula

    {{{x[1,2]}}} = {{{(2 +- sqrt(184))/(2*3)}}} = {{{(2 +- sqrt(184))/6}}} = {{{(1 +- sqrt(46))/3}}}.


The roots are  {{{x[1]}}} = {{{(1 - sqrt(46))/3}}} = -1.92744,

and            {{{x[2]}}} = {{{(1 + sqrt(46))/3}}} = 2.59411  (approximately).


The root {{{x[1]}}} is not in the equation's domain - so, we reject it.

The root {{{x[2]}}}  is in the domain, so we accept it, and this root is the unique solution to equation #1.
</pre>

Equation &nbsp;#1 &nbsp;is solved.


You can check it on your own that my solution &nbsp;x = 2.59411  &nbsp;is correct, &nbsp;by substituting it into the equation.

I did it and obtained a perfect match.