Subtraction of Logarithms: Inequality

To solve the inequality involving logarithms with base $\frac{1}{2}$, we will perform the following steps:

• Step 1: Apply the subtraction property of logarithms.
• Step 2: Simplify and solve the inequality.

Let's go through the steps:

Step 1: Simplify both sides using the logarithm subtraction rule:

Left side: $\log_{\frac{1}{2}}5 - \log_{\frac{1}{2}}4 = \log_{\frac{1}{2}}\left(\frac{5}{4}\right)$

Right side: $\log_{\frac{1}{2}}x - \log_{\frac{1}{2}}3 = \log_{\frac{1}{2}}\left(\frac{x}{3}\right)$

This gives us the inequality:

$\log_{\frac{1}{2}}\left(\frac{5}{4}\right) \le \log_{\frac{1}{2}}\left(\frac{x}{3}\right)$

Step 2: Since $\frac{1}{2}$ is less than 1, the inequality sign flips when we remove the logarithms.

This gives: $\frac{5}{4} \ge \frac{x}{3}$

Multiplying both sides by 3 to solve for $x$:

$3 \cdot \frac{5}{4} \ge x$

$\frac{15}{4} \ge x$

Thus, $x \le \frac{15}{4}$, which simplifies to $x \le 3.75$.

Since we assumed $x > 0$, the final solution is:

$0 < x \le 3.75$

To solve this problem, we'll apply the properties of logarithms and inequality manipulation.

Initially, consider the given inequality:

$\log_2 3 - \log_2 (x + 3) \le 8$

Using the quotient rule of logarithms, combine the logs:

$\log_2 \left(\frac{3}{x + 3}\right) \le 8$

The inequality $\log_2 \left(\frac{3}{x + 3}\right) \le 8$ can be rewritten by converting the logarithm to an exponential form:

$\frac{3}{x + 3} \le 2^8$

Since $2^8 = 256$, substitute to get:

$\frac{3}{x + 3} \le 256$

To remove the fraction, multiply both sides by $x + 3$, assuming $x + 3 > 0$ to maintain the inequality direction:

$3 \le 256(x + 3)$

Divide by 256 to isolate $x+3$:

$\frac{3}{256} \le x + 3$

Subtract 3 from both sides to solve for $x$:

$x \ge \frac{3}{256} - 3$

Given the problem's constraints about the positivity of the logarithm's argument, ensure $x > -3$. Our derived inequality starts from $x \ge \frac{3}{256} - 3$, which satisfies this, thus correctly addressing the domain assumptions.

In conclusion, the solution to the inequality is:

$x \ge \frac{3}{256} - 3$

To solve this problem, we'll follow these steps:

• Step 1: Simplify the logarithmic expressions using $\log_b A - \log_b B = \log_b \left(\frac{A}{B}\right)$.
• Step 2: Compare the simplified expressions' arguments.
• Step 3: Solve the algebraic inequality involving the arguments.

Now, let's work through each step:

Step 1: Use the property $\log_{13} A - \log_{13} B = \log_{13}\left(\frac{A}{B}\right)$ to simplify:

$\log_{13}(2x^2+3) - \log_{13}2 = \log_{13}\left(\frac{2x^2+3}{2}\right)$

$\log_{13}7 - \log_{13}x^2 = \log_{13}\left(\frac{7}{x^2}\right)$

Step 2: This gives the inequality:

$\log_{13}\left(\frac{2x^2+3}{2}\right) \le \log_{13}\left(\frac{7}{x^2}\right)$

Since the logarithm function is monotonically increasing, we can drop the logs and solve:

$\frac{2x^2+3}{2} \le \frac{7}{x^2}$

Multiplying through by $2x^2$, to eliminate fractions, ensures none of the values of $x$ is zero, which would cause division by zero:

$(2x^2 + 3)x^2 \le 14$

Expanding gives a quadratic inequality:

$2x^4 + 3x^2 - 14 \le 0$

Step 3: Substitute $u = x^2$ to transform into quadratic form:

$2u^2 + 3u - 14 \le 0$

Find the critical points by solving the equation $2u^2 + 3u - 14 = 0$:

$u = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 2 \cdot (-14)}}{2 \cdot 2}$

$u = \frac{-3 \pm \sqrt{9 + 112}}{4}$

$u = \frac{-3 \pm \sqrt{121}}{4}$

$u = \frac{-3 \pm 11}{4}$

This gives the roots $u = 2$ and $u = -\frac{7}{2}$. Only non-negative values for $u$ make sense since $u = x^2$, so consider:

$x^2 = u \le 2$

Thus, $-\sqrt{2} \le x \le \sqrt{2}$.

Therefore, the solution to the problem is $-\sqrt{2} \le x \le \sqrt{2}$.