Question 722699
There may be a shorter, more clever way to do this but here goes...<br>
As you will soon see, we will be interested in when the exponent is positive and when it will be negative. (A zero exponent results in a 1 which is not greater than 1. So we factor it:
{{{x^2-18x-56 = (x-4)(x-14)}}}
From this we can tell that...
If (x < 4 or x > 14), then {{{x^2-18x-56}}} will be positive. 
and
if (x > 4 and x < 14), then {{{x^2-18x-56}}} will be negative.
(If you can't see this, then try some of these x's out.)<br>
Now let's start the solution. We will break it down into cases that cover all possible values for the logarithm. (Note: To save space I am omitting the solution to some simple logarithmic inequalities from the main discussion of the solution. If you need to see these solutions, they are at the end.)<ol><li>First of all, the argument of any log may not be zero or negative. In other words it must be positive. If x/6 > 0 then
x > 0</li><li>{{{log(2, (x/6)) > 1}}} If this is true then any positive exponent will make the left side greater than 1. So we need to solve:
{{{log(2, (x/6)) > 1}}} and exponent is positive. Or, using what we found earlier about the exponents:
{{{log(2, (x/6)) > 1}}} and (x < 4 or x > 14)
Solving the log inequality we get:
x > 12 and (x < 4 or x > 14)
This resolves to just
x > 14</li><li>{{{log(2, (x/6)) < -1}}} If this is true then the absolute value will be greater than 1. And again we will need a positive exponent. So:
{{{log(2, (x/6)) < -1}}} and (x < 4 or x > 14)
x < 3 and (x < 4 or x > 14)
which resolves to:
x < 3</li><li>{{{0 < log(2, (x/6)) < 1}}} If this is true then the log (and its absolute value) is a positive, proper fraction. To make this greater than 1 we will need a negative exponent. So
{{{0 < log(2, (x/6)) < 1}}} and (x > 4 and x < 14)
Solving...
x > 6 and x < 12 and (x > 4 and x < 14)
which resolves to:
x > 6 and x < 12</li><li>{{{-1 < log(2, (x/6)) < 0}}} If this is true then the log is a negative proper fraction but its absolute value will be a positive, proper fraction. To make this greater than 1 we will again need a negative exponent. So
{{{-1 < log(2, (x/6)) < 0}}} and (x > 4 and x < 14)
Solving...
x > 3 and x < 6 and (x > 4 and x < 14)
which resolves to:
x > 4 and x < 6</li><li>The only possible values for the logarithm that we have not considered are 0, 1 and -1. If the log is equal to any of these numbers then the left side cannot possible be greater than zero because:<ul><li>If the log is zero, zero to any power cannot be greater than zero.</li><li>If the log is 1 or -1, its absolute value will be a 1 and 1 to any power cannot be greater than one.</li></ul>So we get no additional solutions with the log is 0, 1 or -1.</li></ol>Putting this all together we get:
(0 < x < 3) [from cases 1 and 3]
or
(4 < x < 6) [from case 5]
or
(6 < x < 12) [from case 4]
or
(x > 14) [from case 2]
IOW: The solutions to the inequality are all positive numbers <u>except</u> 3, 4, 6, 12, 14, any number between 3 and 4 or any number between 12 and 14.<br>
Solutions to the logarithmic inequalities:
{{{log(2, (x/6)) > 1}}}
In exponential form:
{{{x/6 > 2^1}}}
{{{x/6 > 2}}}
{{{x > 12}}}<br>
{{{log(2, (x/6)) < -1}}}
In exponential form:
{{{x/6 < 2^(-1)}}}
{{{x/6 > 1/2}}}
{{{x > 3}}}<br>
{{{0 < log(2, (x/6)) < 1}}}
In exponential form:
{{{2^0 < x/6 < 2^1}}}
{{{1 < x/6 < 2}}}
{{{6 < x < 12}}}<br>
{{{-1 < log(2, (x/6)) < 0}}}
In exponential form:
{{{2^(-1) < x/6 < 2^0}}}
{{{1/2 < x/6 < 1}}}
{{{3 < x < 6}}}<br>