Question 634548
{{{log(x, (5)) < log(5, (x))}}}
You are right in thinking we need a common base, either x or 5 or some third number. Using some third number will work but it makes things more complicated (as you will see shortly), so we will use a common base of x or 5. I'll use x. Using the base conversion formula, {{{log(a, (p)) = log(b, (p))/log(b, (a))}}} to convert the base x log into an expression of base 5 logs:
{{{log(5, (5))/log(5, (x)) < log(5, (x))}}}
The numerator of the fraction is equal to 1 (since {{{5 = 5^1}}}) so this will simplify. (This is why using x or 5 as the common base makes things simpler.):
{{{1/log(5, (x)) < log(5, (x))}}}<br>
As you indicated we could continue by making one side zero, etc. And I will do this shortly. But this inequality can be solved much more easily if you look at what it is saying and if you understand certain things about how numbers work.<br>
This equality says that some number, {{{log(5, (x))}}}, is greater than its reciprocal (1/q is the reciprocal of q). When can a number be less than its reciprocal? If you understand reciprocals well you will know that<ul><li>If {{{log(5, (x))}}} is positive (i.e. {{{log(5, (x)) > 0}}}, then {{{log(5, (x))}}} must be greater than 1. "Positive  and  greater than 1" translates into:
{{{log(5, (x)) > 0}}} and {{{log(5, (x)) > 1}}}
which simplifies to just:
{{{log(5, (x)) > 1}}}</li><li>If {{{log(5, (x))}}} is negative (i.e. {{{log(5, (x)) < 0}}}, then it works the opposite of the way it works for positives. We want {{{log(5, (x))}}} to be between 0 and -1. (Think about this and think about reciprocals. It should make sense.)  "Negative and between 0 and -1" translates into:
{{{log(5, (x)) < 0}}} and {{{log(5, (x)) < 0}}} and {{{log(5, (x)) > -1}}}
which simplifies to just:
{{{log(5, (x)) < 0}}} and {{{log(5, (x)) > -1 }}}</li></ul>Since {{{log(5, (x))}}} could be positive or negative, the full solution will come from:
({{{log(5, (x)) > 1}}}) or ({{{log(5, (x)) < 0}}} and {{{log(5, (x)) > -1 }}})
If this makes sense to you and if you think your teacher will accept this, then skip to "THE FINISH" below where I will finish this. Otherwise, keep reading. I will now solve this by getting one side to be zero, etc.<br>
{{{1/log(5, (x)) < log(5, (x))}}}
First let's get rid of the fraction. To do this we will multiply each side by {{{log(5, (x))}}}. But we have to careful doing this because this is an inequality, {{{log(5, (x))}}} might be negative and if we multiply an inequality by a negative number we must reverse the inequality. To do this correctly we must generate two inequalities, one for when {{{log(5, (x))}}} is positive (and we don't reverse the inequality) and one for when {{{log(5, (x))}}} is negative (and we do reverse the inequality):
({{{log(5, (x)) > 0}}} and {{{1 < (log(5, (x)))^2}}}) or ({{{log(5, (x)) < 0}}} and {{{1 > (log(5, (x)))^2}}})
Making the right side 2nd and 4th inequalities a zero:
({{{log(5, (x)) > 0}}} and {{{0 < (log(5, (x)))^2 - 1}}}) or ({{{log(5, (x)) < 0}}} and {{{0 > (log(5, (x)))^2 - 1}}})
Factoring the 2nd and 4th inequalities (using the difference of squares pattern: {{{a^2-b^2 = (a+b)(a-b)}}}):
({{{log(5, (x)) > 0}}} and {{{0 < (log(5, (x)) + 1)(log(5, (x)) - 1)}}}) or ({{{log(5, (x)) < 0}}} and {{{0 > (log(5, (x)) + 1)(log(5, (x)) - 1)}}})<br>
Next I will deal with this in parts. The first "half":
({{{log(5, (x)) > 0}}} and {{{0 < (log(5, (x)) + 1)(log(5, (x)) - 1)}}})
says that the log is positive and a product of two factors is greater than zero (IOW, positive). (Note: Always read inequalities form where the variable is. The variable in {{{0 < (log(5, (x)) + 1)(log(5, (x)) - 1)}}} is on the right so we should read it from right to left. And from right to left, "<" is a "greater than".) And how does a product of two numbers become positive? Answer: If both numbers are positive or if both numbers are negative. And since {{{log(5, (x))}}} is positive, doesn't {{{(log(5, (x)) + 1)}}} have to be positive, too? So we can't have two negative, jsut two positives And since we already know that {{{(log(5, (x)) + 1)}}} is positive, then we just have to say that {{{(log(5, (x)) - 1)}}} is positive, too. negative! So the first half simplifies to:
({{{log(5, (x)) > 0}}} and {{{(log(5, (x)) - 1) > 0}}})
Adding 1 to both sides of the second inequality we get:
({{{log(5, (x)) > 0}}} and {{{log(5, (x)) > 1}}})
which simplifies to just:
({{{log(5, (x)) > 1}}})<br>
Next, let's look at the second "half":
({{{log(5, (x)) < 0}}} and {{{0 > (log(5, (x)) + 1)(log(5, (x)) - 1)}}})
which says that the log is negative and a product of two numbers is negative. A product of two numbers is negative only when one is positive and the other is negative. Since the log is negative, the factor {{{(log(5, (x)) - 1)}}} must be negative, too. So the {{{(log(5, (x)) + 1)}}} must be positive:
({{{log(5, (x)) < 0}}} and {{{(log(5, (x)) + 1) > 0}}})
Subtracting 1 from both sides of the second inequality we get:
({{{log(5, (x)) < 0}}} and {{{log(5, (x)) > -1}}})<br>
Putting our simplified "halves" back together we have:
({{{log(5, (x)) > 1}}}) or ({{{log(5, (x)) < 0}}} and {{{log(5, (x)) > -1 }}})
Note how this is exactly what we got using some logic and our understanding of reciprocals!<br>
THE FINISH<br>
Solving
({{{log(5, (x)) > 1}}}) or ({{{log(5, (x)) < 0}}} and {{{log(5, (x)) > -1 }}})
Rewriting each of these in exponential form:
({{{x > 5^1}}}) or ({{{x < 5^0}}} and {{{x > 5^(-1)}}})
Simplifying:
({{{x > 5}}}) or ({{{x < 1}}} and {{{x > 1/5}}})
This is our solution. Translated: x can be any number larger than 5 or any number between 1/5 and 1.