Value of c in equation $1+\log_2(2x^2+2x+\frac{7}{2}) \ge \log_2(cx^2+c)$ for at-least one solution

Question

Find all the values of the parameters c for which the following inequality has at least one solution. $1+\log_2(2x^2+2x+\frac{7}{2}) \ge \log_2(cx^2+c)$

I will elaborate the following steps

1.$\log_2(4x^2+4x+7) \ge \log_2(cx^2+c)$

2. Converting it into ln form or base e form

3.$\ln(4x^2+4x+7) \le \ln(cx^2+c)$ [Change of sign occurs]

4.$(4x^2+4x+7) \le (cx^2+c)$

5.$(4-c)x^2+4x+7-c \le 0$

6.Let y=$(4-c)x^2+4x+7-c$

7.$4^2-4(4-c)(7-c)\ge0$ for roots to exist

8. Solving we get $3\le c \le8$ but answer is $0< c \le8$

I cannot find my mistake .

Answer 1

As noted in the comments, converting from base $2$ to base $e$ doesn't cause the direction of the inequality to reverse.

In fact, there there's no real need to convert to base $e$, since $\log_2(x)$ is also an increasing function.

To find the qualifying values of $c$, we can argue as follows . . .

First note that we must have $c > 0$, else $\log_2(cx^2+c)$ would be undefined for all $x\in\mathbb{R}$.

Suppose $c > 0$ is such that $1+\log_2(2x^2+2x+\frac{7}{2}) \ge \log_2(cx^2+c)$, for some $x\in \mathbb{R}$. \begin{align*} \text{Then}\;\;&1+\log_2(2x^2+2x+{\small{\frac{7}{2}}}) \ge \log_2(cx^2+c)\\[4pt] \iff\;&\log_2(4x^2+4x+7) \ge \log_2(cx^2+c)\\[4pt] \iff\;&4x^2+4x+7 \ge cx^2+c\\[4pt] \iff\;&(4-c)x^2+4x+(7-c) \ge 0\\[4pt] \end{align*} Let $f(x)=(4-c)x^2+4x+(7-c)$.

Consider three cases . . .

If $c < 4$, then $f$ is a quadratic function with positive leading coefficient, which assumes nonnegative values, hence $c\;$is qualifying for $0 < c < 4$. Note:$\;$In your analysis, you missed this case.

If $c = 4$, then $f(x)=4x+3$ which assumes nonnegative values, hence $c=4$ is qualifying.

If $c > 4$, then $f$ is a quadratic function with negative leading coefficient, so $f$ assumes nonnegative values if and only if $f$ has a real root. As in your analysis, this amounts to solving the inequality $-4(c-3)(c-8)\ge 0$, hence, incorporating the restriction $c > 4$ (since that's the case we're in), $c\;$is qualifying for $4 < c \le 8$.

Combining the results of the three cases, we get that $c\;$is qualifying if and only if $0 < c \le 8$.

Answer 2

In step 2., there is no need to convert the base. More importantly, you should not flip the inequality sign, as $\ln 2>0$.

So in step 4., we should have $4x^2+4x+7\geq cx^2+cx$. Also do not forget about the extra condition that we lose: you need to determine $c$ so that at least one of the solutions satisfies $cx^2+cx>0$, otherwise you cannot substitute into the original expressions. (The transformations are only equivalent if all the expresssions make sense.)

One further problem: you cannot use the discriminant if the equation is not really quadratic, so you have to check the case $c=4$ separately.

But I think you are on the right track. Fix the above problems, and finish the proof, it is going to be fine.