<PRE><B>Question 730965</B>
Remember that log(base n) (x) + log(base n) (y) = log(base n) (xy)&lt;P&gt;
LogBase4(x-2)+LogBase4(x+10)=3&lt;P&gt;
logbase((x-2)*(x+10)) = 3&lt;P&gt;
logbase4(x^2 + 8x -20) = 3&lt;P&gt;
Recall that logbasen (x) = y  if and only if n^y = x. &lt;P&gt;
{{{x^2 + 8x - 20 = 4^3 = 64}}}&lt;P&gt;
{{{x^2 + 8x - 84 = 0}}}
(x+14)(x-6) = 0&lt;P&gt;
x=-14 and x=6&lt;P&gt;
The log of a negative number is imaginary.  So only x=6 works.  -14 results in logbase4 (-16) + logbase4 (-4) = 3 .  The log of negative numbers isn&#39;t real.
&lt;P&gt;Hope the solution helped.  Sometimes you need more than a solution.  Contact fcabanski@hotmail.com for online, private tutoring, or personalized problem solving (quick for groups of problems.)
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