Question 1198977

solve for x: 1+log2(x^2-4x-16)=log2(x^2-3x+4)
<pre>You DON'T need to use the quadratic equation formula to solve this, as that woman did!

{{{matrix(2,3, 1 + log (2, (x^2 - 4x - 16)), "=", log (2, (x^2 - 3x + 4)), 1, "=", log (2, (x^2 -3x + 4)) - log (2, (x^2 - 4x - 16)))}}}
                  {{{matrix(1,3, 1, "=", log (2, ((x^2 - 3x + 4)/(x^2 - 4x - 16))))}}} ----- Applying {{{matrix(1,4, log (a, (b)) - log (a, c), "=", "=====>", log (a, (b/c)))}}}
                  {{{matrix(2,3, (x^2 - 3x + 4)/(x^2 - 4x - 16), "=", 2^1, (x^2 - 3x + 4)/(x^2 - 4x - 16), "=", 2)))}}} ----- Applying {{{matrix(1,7, log (a, (b)), "=", c, "=====>", b, "=", a^c)}}}
                {{{matrix(1,3, 2(x^2 - 4x - 16), "=", x^2 - 3x + 4)}}} ----- Cross-multiplying
        {{{matrix(3,3, 2x^2 - 8x - 32, "=", x^2 - 3x + 4, 2x^2 - x^2 - 8x + 3x - 32 - 4, "=", 0, x^2 - 5x - 36, "=", 0)}}}
                (x - 9)(x + 4) = 0
                         x - 9 = 0     or      x + 4 = 0

                         {{{highlight_green(system(matrix(1,3, x, "=", 9), or, matrix(1,3, x, "=", - 4)))}}}</pre>