Question

Log(2x²+√2)to the base 2=√(x²+1) / (x²+2)

Answer 1

0

Answer 2

To solve the equation $\log_2(2x^2+\sqrt{2}) = \frac{\sqrt{x^2+1}}{x^2+2}$, we can analyze the range of the expressions on both sides of the equation.

Let the Left Hand Side (LHS) be $f(x) = \log_2(2x^2+\sqrt{2})$ and the Right Hand Side (RHS) be $g(x) = \frac{\sqrt{x^2+1}}{x^2+2}$.

Analysis of the LHS, $f(x)$:

1. Domain: The argument of the logarithm must be positive. $2x^2+\sqrt{2} > 0$. Since $x^2 \ge 0$, $2x^2 \ge 0$, so $2x^2+\sqrt{2} \ge \sqrt{2}$. Thus, the argument is always positive for all real $x$. The domain of $f(x)$ is $x \in \mathbb{R}$.

2. Minimum Value: The expression $2x^2+\sqrt{2}$ has its minimum value when $x^2=0$, i.e., $x=0$. Minimum value of $2x^2+\sqrt{2} = 2(0)^2+\sqrt{2} = \sqrt{2}$.

3. Range of $f(x)$: Since $\log_2(t)$ is an increasing function, the minimum value of $f(x)$ occurs at $x=0$. $f(0) = \log_2(\sqrt{2}) = \log_2(2^{1/2}) = \frac{1}{2}$. As $|x| \to \infty$, $2x^2+\sqrt{2} \to \infty$, so $\log_2(2x^2+\sqrt{2}) \to \infty$. Therefore, the range of $f(x)$ is $[1/2, \infty)$. So, $f(x) \ge 1/2$.

Analysis of the RHS, $g(x)$:

1. Domain: The expression under the square root must be non-negative. $x^2+1 \ge 0$. This is true for all real $x$. The denominator $x^2+2$ is always positive. The domain of $g(x)$ is $x \in \mathbb{R}$.

2. Substitution: Let $u = x^2$. Since $x^2 \ge 0$, $u \ge 0$. The expression becomes $h(u) = \frac{\sqrt{u+1}}{u+2}$ for $u \ge 0$.

3. Maximum Value: To find the range of $h(u)$, we can evaluate it at $u=0$ and as $u \to \infty$, and check its derivative. At $u=0$ (i.e., $x=0$): $h(0) = \frac{\sqrt{0+1}}{0+2} = \frac{1}{2}$. As $u \to \infty$: $h(u) = \frac{\sqrt{u+1}}{u+2} \approx \frac{\sqrt{u}}{u} = \frac{1}{\sqrt{u}} \to 0$. Let's find the derivative of $h(u)$: $h'(u) = \frac{\frac{1}{2\sqrt{u+1}}(u+2) - \sqrt{u+1}(1)}{(u+2)^2} = \frac{(u+2) - 2(u+1)}{2\sqrt{u+1}(u+2)^2} = \frac{u+2-2u-2}{2\sqrt{u+1}(u+2)^2} = \frac{-u}{2\sqrt{u+1}(u+2)^2}$. For $u > 0$, $h'(u) < 0$, which means $h(u)$ is a strictly decreasing function for $u \ge 0$.

4. Range of $g(x)$: Since $h(u)$ is decreasing for $u \ge 0$, its maximum value occurs at $u=0$. Therefore, the maximum value of $g(x)$ is $1/2$ (at $x=0$). The range of $g(x)$ is $(0, 1/2]$. So, $g(x) \le 1/2$.

Conclusion:

We have established that:

1. $f(x) = \log_2(2x^2+\sqrt{2}) \ge 1/2$ for all $x \in \mathbb{R}$.
2. $g(x) = \frac{\sqrt{x^2+1}}{x^2+2} \le 1/2$ for all $x \in \mathbb{R}$.

For the equation $f(x) = g(x)$ to hold, both sides must be equal to $1/2$. This means we need to find $x$ such that: $f(x) = 1/2 \implies \log_2(2x^2+\sqrt{2}) = 1/2 \implies 2x^2+\sqrt{2} = 2^{1/2} = \sqrt{2}$ $2x^2 = 0 \implies x^2 = 0 \implies x = 0$.

And also: $g(x) = 1/2 \implies \frac{\sqrt{x^2+1}}{x^2+2} = 1/2 \implies 2\sqrt{x^2+1} = x^2+2$ Squaring both sides: $4(x^2+1) = (x^2+2)^2$ $4x^2+4 = x^4+4x^2+4$ $x^4 = 0 \implies x = 0$.

Since both conditions are satisfied only when $x=0$, the unique solution to the equation is $x=0$.