Question 933453
<pre>
Log(base 2) (x+14)+log(base 2)x=5

{{{log (2, (x + 14)) + log (2, (x)) = 5}}}
Smaller variable-expression, x, MUST be > 0. We then get: x MUST be > 0. The problem then becomes:
{{{log (2, (x + 14)) + log (2, (x)) = 5}}}, with x > 0.
       {{{log (2, (x + 14))x = 5}}} ---- Applying {{{log (a, (b)) + log (a, (c))}}} = {{{log (a, (bc))}}}
             {{{x(x + 14) = 2^5}}} --- Converting to EXPONENTIAL form
             {{{x^2 + 14x = 32}}}
         {{{x^2 + 14x - 32 = 0}}}
     (x - 2)(x + 16) = 0 ----- Factoring TRINOMIAL
      x - 2 = 0      OR      x + 16 = 0 --- Setting each factor = to 0
         <font color = red><font size = 4><b>x = 2</font></font></b>      OR           x = - 16 <font color = red><font size = 4><b>(IGNORE)</font></font></b>.

x = 2 is the ONLY solution, since the other value, - 16, is NOT > 0.</pre>