<PRE><B>Question 727931</B>
&lt;pre&gt;
There are TWO solutions for x, not just one.  To do this you
need to know the change of base formula {{{log(b,a) = log(c,a)/log(c,b)}}}

and the definition of a logarithm:  {{{log(b,a)=c}}} where c is the exponent to
which base b must be raised to give a.  That is {{{log(b,a)=c}}} is equivalent
to the equation {{{a=b^c}}}  


log&lt;sub&gt;3&lt;/sub&gt;x + log&lt;sub&gt;x&lt;/sub&gt;9 = 3

Use the change of base formula on the second term:

log&lt;sub&gt;3&lt;/sub&gt;x + {{{log(3,9)/log(3,x)}}} = 3

Multiply through by log&lt;sub&gt;3&lt;/sub&gt;x

(log&lt;sub&gt;3&lt;/sub&gt;x)² + log&lt;sub&gt;3&lt;/sub&gt;9 = 3·log&lt;sub&gt;3&lt;/sub&gt;x

log&lt;sub&gt;3&lt;/sub&gt;9 = 2 since the exponent to which the base 3 must be
raised to give 9 is the exponent 2, since 3² = 9 

(log&lt;sub&gt;3&lt;/sub&gt;x)² + 2 = 3·log&lt;sub&gt;3&lt;/sub&gt;x

(log&lt;sub&gt;3&lt;/sub&gt;x)² - 3·log&lt;sub&gt;3&lt;/sub&gt;x + 2 = 0

That factors as a quadratic in log&lt;sub&gt;3&lt;/sub&gt;x

(log&lt;sub&gt;3&lt;/sub&gt;x - 2)(log&lt;sub&gt;3&lt;/sub&gt;x - 1) = 0

Use the zero-factor property:

log&lt;sub&gt;3&lt;/sub&gt;x - 2 = 0;   log&lt;sub&gt;3&lt;/sub&gt;x - 1 = 0;
    log&lt;sub&gt;3&lt;/sub&gt;x = 2;       log&lt;sub&gt;3&lt;/sub&gt;x = 1;
        x = 3&lt;sup&gt;2&lt;/sup&gt;          x = 3&lt;sup&gt;1&lt;/sup&gt;
        x = 9           x = 3

Two solutions: 9 and 3.

Checking x = 9

log&lt;sub&gt;3&lt;/sub&gt;x + log&lt;sub&gt;x&lt;/sub&gt;9 = 3
log&lt;sub&gt;3&lt;/sub&gt;9 + log&lt;sub&gt;9&lt;/sub&gt;9 = 3
  2 + 1 = 3
      3 = 3

Checking x = 3

log&lt;sub&gt;3&lt;/sub&gt;x + log&lt;sub&gt;x&lt;/sub&gt;9 = 3
log&lt;sub&gt;3&lt;/sub&gt;3 + log&lt;sub&gt;3&lt;/sub&gt;9 = 3
  1 + 2 = 3
      3 = 3

Both answers check.

Edwin&lt;/pre&gt;
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