Question 1019829
<pre>
{{{log(1/3,(x^2+3x+4)) <= -1}}}

Let the left side = y

{{{log(1/3,(x^2+3x+4)) = y}}}

{{{(1/3)^y = x^2+3x+4}}}

Take the log base 10 of both sides

{{{log((1/3)^y) = log((x^2+3x+4))}}}

{{{y*log((1/3)) = log((x^2+3x+4))}}}

{{{y = log((x^2+3x+4))/log((1/3))}}}

The original inequality was

{{{log(1/3,(x^2+3x+4)) <= -1}}}

So {{{y<=-1}}}.  Substituting for y

{{{log((x^2+3x+4))/log((1/3))<=-1}}}

We know that {{{log((1/3))=-.4771}}} is a negative 
number, so when we multiply both sides
of the inequality by it, the inequality
symbol will reverse:

{{{log((x^2+3x+4))>=-1*log((1/3))}}}

Use a principle of logarithms to rewrite
the right side:

{{{log((x^2+3x+4))>=log(((1/3)^(-1)))}}}

Since raising to the -1 power is the same
as taking the rciprocal, the above becomes:

{{{log((x^2+3x+4))>=log((3))}}}

Since logs base 10 are in ascending order,

{{{x^2+3x+4>=3}}}

{{{x^2+3x+1>=0}}}

That has critical numbers

 {{{x = (-3 +- sqrt( 3^2-4*1*1 ))/(2*1) }}}

 {{{x = (-3 +- sqrt(5))/2 }}}

Put those two values on a number line and
choose a test point in each region to discover that
the solution set is

 {{{x <= (-3 - sqrt(5))/2 }}} or {{{x >=(-3 + sqrt(5))/2 }}}

Edwin</pre>