Question 1194878
Please help me solve this question:log2{√(x³-2x²+x)}=1+log2(x-1)
<pre>{{{matrix(3,3, log (2, (sqrt(x^3  -  2x^2 + x))), "=", 1 + log (2, (x  -  1)), 
log (2, (x^3  -  2x^2 + x)^(1/2)), "=", 1 + log (2, (x  -  1)), 
(1/2) * log (2, (x^3  -  2x^2 + x)), "=", 1 + log (2, (x  -  1)))}}}
{{{matrix(1,3, log (2, (x^3  -  2x^2 + x)), "=", 2 + 2 * log (2, (x  -  1)))}}} ------ Multiplying by LCD, 2 
{{{matrix(1,3, log (2, (x^3  -  2x^2 + x))  -  2 * log (2, (x  -  1)), "=", 2)}}} ------ Subtracting {{{2 * log (2, (x  -  1))}}} from both sides
{{{matrix(4,3, log (2, (x^3  -  2x^2 + x))  -  log (2, (x  -  1)^2), "=", 2,  
log (2, ((x^3  -  2x^2 + x)/(x  -  1)^2)), "=", 2,
log (2, ((x(x^2  -  2x + 1))/(x  -  1)^2)), "=", 2, 
log (2, (x(x - 1)^2/(x - 1)^2)), "=", 2)}}}
                    {{{matrix(1,3, (x(x  -  1)^2)/(x  -  1)^2, "=", 2^2)}}} ------ Converting to EXPONENTIAL form 
At this point, with x - 1 in the denominator, the equation's "x" values can be ANY REAL NUMBER, EXCEPT 1.
                      {{{matrix(1,3, x(x - 1)^2, "=", 4(x - 1)^2)}}} ------- Cross-multiplying
            {{{matrix(3,3, x(x - 1)^2 - 4(x - 1)^2, "=", 0, (x - 4)(x - 1)^2, "=", 0, (x - 4)(x - 1)(x - 1), "=", 0)}}}
Therefore, x = 4 and also, x = 1 (DOUBLE-ROOT)
However, since x CANNOT = 1 ({{{x <> 1}}}, as stated above), the ONLY solution is: <font size = 5><font color = red><b>x = 4</font></font></b>. 

Just to make sure that no errors were made, and since this involves LOGARITHMS and square roots, you'll need to CHECK to make sure that x = 4 is 
NOT an EXTRANEOUS solution.</pre>