Please evaluate my solutions.
Find all values of the parameter $\alpha\in R $ for which the following inequality is valid for all $x \in R$ $$1+\log _5\left(x^2+1\right)\ge \log _5\left(ax^2+4x+a\right)$$
The range of values can be expressed in the form of $(A,B]$
I have gathered 2 ways of finding the solution:
1)Change log bases
$$\log_5(5) + \log_5\frac{x^2+1}{ax^2+4x+a} >= 0$$
$$\log_5\frac{5x^2+5}{ax^2+4x+a} >= 0$$
$$\log_\frac{5x^2+5}{ax^2+4x+a}{5} >=0$$
$$\frac{5x^2+5}{ax^2+4x+a} >0$$
Got two inequalities
$$ax^2+4x+a > 0 ... (1)$$
$$5x^2+5 >0 ... (2)$$
$$ax^2+4x+a > 0 ... (1)$$
Find the discriminant:
$$a=2,a=-2$$
$$5x^2+5 >0 ...(2)$$
$$5x^2>-5$$
$$x>(-1)^\frac{1}{2}$$
My answer is $(2, +\infty)$
2) solve by using the same base.
$$\log_5(5) + \log_5(x^2+1) >= \log_5(ax^2+4x+a)$$
$$\log_5(5x^2+5) >= \log_5(ax^2+4x+a)$$
Got 3 inequalities:
$$5x^2+5>=ax^2+4x+a ... (1)$$
$$5x^2+5>=0 ... (2)$$
$$ax^2+4x+a >= 0 ... (3)$$
$$(5-a)x^2-4x+(5-a)>=0 ...(1)$$
Find the discriminant:
$$16^2-4(5-a)^2$$
$$a = 3, a=7$$
$$5x^2+5 >0 ... (2)$$
$$5x^2>-5$$
$$x>(-1)^\frac{1}{2}$$
$$ax^2+4x+a >= 0 ... (3)$$
Find the discriminant:
$$a=2,a=-2$$
My answer is $$(2,7)$$
I know both answers are wrong because the correct one is $$(2,3]$$
Can we get the correct answer from both ways at the top?
Answer
Your first approach is much more complicated than necessary, and there are a couple of mistakes.
To rephrase the original problem slightly,
you were to find $A$ and $B$ such that $A < a \leq B$ is a necessary and sufficient condition for the following inequality to be true:
$$1+\log _5\left(x^2+1\right)\ge \log _5\left(ax^2+4x+a\right). \tag P$$
It is correct that
$$\log_5\frac{5x^2+5}{ax^2+4x+a} \geq 0 \tag Q$$
is a necessary and sufficient condition for (P) to be true.
Since $\log_5 y = \log_y 5$ when both sides of the equation are defined,
your next statement,
$$\log_{\frac{5x^2+5}{ax^2+4x+a}} 5 \geq 0, \tag R$$
is a sufficient condition, but not necessary:
in the case where $\log_5\frac{5x^2+5}{ax^2+4x+a} = 0,$
the expression $\log_{\frac{5x^2+5}{ax^2+4x+a}} 5$ is undefined,
so we cannot say that it is greater than or equal to anything.
In particular, in that case we cannot say that (R) is true,
but nevertheless (Q) and (P) are still true even though (R) is not.
Getting the "$=$" case wrong is not going to hurt you too much in this
question, because here it mainly affects whether you have an open or
closed interval, and you were given which side of the interval is open and
which is closed in the problem statement.
What really hurts you is the next step,
$$\frac{5x^2+5}{ax^2+4x+a} \geq 0,$$
which is a necessary but not sufficient condition for (P) to be true.
Actually, either the fact that you can take a logarithm
of $\frac{5x^2+5}{ax^2+4x+a}$ or that you can use it as the base of
a logarithm tells you immediately that
$$\frac{5x^2+5}{ax^2+4x+a} > 0$$
(notice there is no possibility of "$=$"),
but merely using positive numbers in the base or parameter of your logarithm
does not guarantee that the result is non-negative;
for example, $log_2 \frac12 = log_{(1/2)} 2 = -1.$
A sufficient (and necessary) condition for (Q) and (P) is
$$\frac{5x^2+5}{ax^2+4x+a} \geq 1. \tag S$$
Notice that this is already clear from (Q) without needing to do any
change of base, because $\log_5 y < 0$ whenever $y < 1.$
TL;DR: You should never have attempted to write (R); it's not quite
accurate and it's much more confusing to work with than (Q).
Note that $5x^2+5$ is guaranteed to be positive, since
$x^2\geq 0$ for all $x$ and $5 > 0.$
So (S) implies that $ax^2+4x+a > 0$ (otherwise the ratio would be negative),
and we can multiply both sides of (S) by $ax^2+4x+a.$
Therefore either of your two methods, correctly executed, gives us the
same three inequalities:
\begin{align}
5x^2+5 &\geq ax^2+4x+a \tag 1\\
5x^2+5 &\geq 0 \tag 2\\
ax^2+4x+a &> 0 \tag 3
\end{align}
The statement that all three of these are true for all $x$
is a necessary and sufficient condition for (P).
Notice that I have written ($3$) in a stronger form than you did.
(It says $>$ rather than $\geq$.)
This only matters if you need to find out whether each end of
the interval of your solution set is open or closed.
As you (sort of) observed, if we take the $=$ case of ($1$) we get a quadratic equation,
$$(5-a)x^2-4x+(5-a) = 0, \tag 4$$
whose discriminant is $16^2-4(5-a)^2.$
But you did not say why that matters.
The reason why it matters is that if the discriminant is positive,
then ($4$) has solutions for two distinct values of $x,$
and therefore $(5-a)x^2-4x+(5-a)$ is negative for some values of $x.$
That is, it would not be true that $5x^2+5 \geq ax^2+4x+a$ for all $x.$
But simply finding where the discriminant is zero doesn't tell us much
about whether $5x^2+5 \geq ax^2+4x+a$ for all $x.$
In order to know whether ($1$) will be true, we want to know
the value of $a$ for which the discriminant is not positive,
and we also need $5-a > 0,$ since otherwise very large positive or negative
values of $x$ will make $5x^2+5 < ax^2+4x+a.$
In other words, we need $5-a > 0$ and we need $4(5-a)^2 \geq 16^2.$
Simplifying, $5-a > 0$ and $(5-a)^2 \geq 64,$
therefore $5-a \geq 8,$ so $a \leq 3.$
In particular, note that if $3 < a < 7$ then the discriminant
$16^2-4(5-a)^2$ is positive and $(1)$ is false for some values of $x.$
So you must not let any part of the interval $(3,7)$ be part of
your answer.
I do not know why you thought the answer should be based on an inequality
involving $a$ and the number $7,$ but if it was a guess, it was a bad guess.
The inequality ($2$) is automatically true for all $x$ for reasons
already given. Please do not write anything like $x > (-1)^{1/2}$;
that inequality is not true for any value of $x$
for the simple reason that $(-1)^{1/2}$ is not a real number.
The inequality ($3$) does indeed correspond to a quadratic whose
discriminant is zero when $a = \pm 2.$
Again, however, you did not say why that matters, and you did not say
why you concluded that $a > 2$ rather than $a < -2$ or
$-2 \leq a \leq 2$ or any other condition on $a.$
In fact we need $a$ to be positive in order for $(3)$ to be true
for large values of $x,$ and we rule out $-2 \leq a \leq 2$
because that makes the discriminant non-negative, which implies that
$ax^2+4x+a = 0$ for at least one value of $x.$
So indeed we do need $a > 2,$ whether or not this was a valid deduction
or a lucky guess on your part.
So altogether, $2 < a \leq 3,$ so $A = 2$ and $B = 3.$
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