Solve log inequalities by changing log bases vs. using the same base

$a \in R$ for which the following inequality is valid for all $x \in R$

we can solve it by

1. Using the same base
2. Change log bases.

The usual way and less complicated way is #1. But out of curiosity I wanted to test for #2.

Let’s begin.

Change log bases

Nope, we can’t do the inequality as $\ge 0$. Eg - $\frac{1}{0.99999} \ge 1$ It will never be 0. Let’s change it to a new equation

$R$ is a denominator of $Q$, thus it must be greater than $0$ and $Q$ shows that the fraction needs to be $\ge 1$. Hence, $\log_\frac{5x^2+5}{ax^2+4x+a}{5} \in(0,1]$.

Let’s take an example and observe.

As you can see $\frac{1}{2}\in (0,1]$

If some number $x$ is $x^{(0,1])} > 0$ then $x \ge 1$

We found the inequalities! As we’ll observe later the inequalities are the same as the method 1 using the same base.

Using the same base

Take from $P$ we can immediately get the inequalities. $5^0=1$ thus $\frac{5x^2+5}{ax^2+4x+a} \ge 1$.

Pheww, boy was that shorter!

Both methods have the same inequalities. Let’s solve them up!

Solve the inequalities

The graph is concave up and has a minimum point because the whole equation is $\ge 0$ therefore $5-a>0$. Find the minimum vertex point

We don’t need to do the equation $2$ because if $a>1$ then $\log_a{(f(x))} \ge \log_a{(g(x))}$ is equivalent to $\begin{cases} f(x) \gt g(x)\\ f(x) \gt 0\\ g(x) \gt 0\\ \end{cases}$

If $g(x) \gt 0$ then automatically $f(x) \gt 0$.

In this case the original equation is $\log_5 5x^2+5 \ge \log_5 ax^2+4x+a$. Clearly $ax^2+4x+a \gt 0$

When $a=3, x=1$ and $a \in [2, 3)$ , $x=imaginary$