Subtraction of Logarithms: Using variables

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

• Simplify the logarithmic expression using logarithmic identities.
• Substitute the simplified result back into the main expression and calculate its value.

Now, let's work through each step:

Step 1: Simplify the logarithmic expression. We'll simplify the parts involving $\log_7$ first, then those involving $\log_2$.

For the terms with $\log_7$:
- Convert $\log_7 x^n$ terms using the power rule: $\log_7 x^7 = 7 \log_7 x$, $\log_7 x^4 = 4 \log_7 x$, and $\log_7 x^3 = 3 \log_7 x$.
- The expression becomes $7 \log_7 x - 4 \log_7 x - 3 \log_7 x$.
- Simple arithmetic yields $0 \log_7 x$, which simplifies to $0$.

For the terms with $\log_2$:
- Similarly, $\log_2 y^n$ terms use the power rule: $\log_2 y^4 = 4 \log_2 y$, $\log_2 y^3 = 3 \log_2 y$, and $\log_2 y = 1 \log_2 y$.
- The expression is $4 \log_2 y - 3 \log_2 y - 1 \log_2 y$.
- Simple arithmetic gives $0 \log_2 y$, which also simplifies to $0$.

Step 2: Substitute these back into the original expression:

Original expression:
$\ln 4 \times (0 + 0) = \ln 4 \times 0 = 0$.

Therefore, the value of the expression is $\textbf{0}$.

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 the problem, we proceed as follows:

Given the equation:

$\ln 8x \times \log_7 e^2 = 2(\log_7 8 + \log_7 x^2 - \log_7 x)$

• Step 1: Express $\ln 8x$ using the change of base formula:

• $\ln 8x = \frac{\log_7 (8x)}{\log_7 e}$

• Step 2: Substitute into the original equation:

• $\frac{\log_7 (8x)}{\log_7 e} \cdot \log_7 e^2 = 2(\log_7 8 + \log_7 x^2 - \log_7 x)$

• Step 3: Simplify using $\log_7 e^2 = 2 \log_7 e$:

• $\frac{\log_7 (8x)}{\log_7 e} \cdot 2 \log_7 e = 2(\log_7 8 + \log_7 x^2 - \log_7 x)$

• Step 4: Cancel $\log_7 e$ and simplify:

• $\log_7 (8x) \cdot 2 = 2(\log_7 8 + \log_7 x^2 - \log_7 x)$

• Step 5: Cancel 2 on both sides:

• $\log_7 (8x) = \log_7 8 + \log_7 x^2 - \log_7 x$

• Step 6: Use the properties of logarithms:

• $\log_7 (8x) = \log_7 8 + \log_7 \frac{x^2}{x}$

• Step 7: Simplify $\log_7 \frac{x^2}{x}$:

• $\log_7 (8x) = \log_7 8 + \log_7 x$

• Step 8: Use properties $\log_b m + \log_b n = \log_b (mn)$:

• $\log_7 (8x) = \log_7 (8x)$

• Step 9: This equality is true for all $x > 0$, considering domain restrictions:

• $\text{For } x > 0$

Thus, the solution is valid for all $x$ such that $x > 0$

Therefore, the correct solution is, For all $\mathbf{x > 0}$.

The equation to solve is $\ln x + \ln(x+1) - \ln 2 = 3$.

Step 1: Combine the logarithms using the product and quotient rules:

$\ln (x(x+1)) - \ln 2 = 3 \quad \text{becomes} \quad \ln \left(\frac{x(x+1)}{2}\right) = 3.$

Step 2: Eliminate the logarithm by exponentiating both sides:

$\frac{x(x+1)}{2} = e^3.$

Step 3: Solve for $x$ by clearing the fraction:

$x(x+1) = 2e^3.$

Step 4: Expand and set up a quadratic equation:

$x^2 + x - 2e^3 = 0.$

Step 5: Use the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = 1$, $b = 1$, and $c = -2e^3$:

$x = \frac{-1 \pm \sqrt{1^2 - 4 \times 1 \times (-2e^3)}}{2 \times 1}.$

Step 6: Simplify under the square root:

$x = \frac{-1 \pm \sqrt{1 + 8e^3}}{2}.$

Step 7: Ensure $x > 0$. Given $\sqrt{1 + 8e^3}$ will be positive, $\frac{-1 + \sqrt{1 + 8e^3}}{2}$ is the valid solution.

Therefore, the solution to the problem is $\frac{-1+\sqrt{1+8e^3}}{2}$.

To solve this logarithmic equation, we will simplify both sides using logarithm properties.

Step 1: Combine the logarithms on the left side.

The left side is $\log_4 9x + \log_4 (x+4) - \log_4 3$. Using the properties of logarithms, we can combine these logs:

$\log_4 \left( \frac{9x(x+4)}{3} \right)$

This simplifies to:

$\log_4 \left(3x(x+4)\right)$

Step 2: Simplify the right side.

The right side is $\ln 2e + \ln \frac{1}{2e}$. Using properties of natural logarithms, combine as follows:

$\ln \left(2e \cdot \frac{1}{2e}\right) = \ln 1 = 0$

Step 3: Equating both sides, we have:

$\log_4 \left(3x(x+4)\right) = 0$

Step 4: Convert the logarithmic equation to an exponential equation. Since the logarithmic expression equals zero, it signifies:

$3x(x+4) = 4^0 = 1$

Step 5: Solve the equation $3x(x+4) = 1$:

Combine and expand the terms:

$3x^2 + 12x - 1 = 0$

Step 6: Solve the quadratic equation using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = 3$, $b = 12$, and $c = -1$:

$x = \frac{-12 \pm \sqrt{12^2 - 4 \times 3 \times (-1)}}{2 \times 3}$

Calculate:

$x = \frac{-12 \pm \sqrt{144 + 12}}{6}$

$x = \frac{-12 \pm \sqrt{156}}{6}$

$x = \frac{-12 \pm \sqrt{4 \times 39}}{6}$

$x = \frac{-12 \pm 2\sqrt{39}}{6}$

$x = \frac{-6 \pm \sqrt{39}}{3}$

Thus, the solution is:

$x = -2 + \frac{\sqrt{39}}{3}$

This matches the correct choice.

Therefore, the solution to the problem is $-2+\frac{\sqrt{39}}{3}$.

To solve this problem, we will follow these steps:

• Step 1: Simplify the left-hand side using logarithm properties.
• Step 2: Simplify the right-hand side using change of base.
• Step 3: Equate simplified forms and solve for $x$.

Now, let's proceed:

Step 1: Simplify the left-hand side:
We can combine the logs as follows:
$\log_5 x + \log_5 (x+2) = \log_5 (x(x+2)) = \log_5 (x^2 + 2x).$
The constants are simplified as:
$\log_2 5 - \log_2 2.5 = \log_2 \left(\frac{5}{2.5}\right) = \log_2 2 = 1.$
Thus, the entire left-hand side becomes:
$\log_5 (x^2 + 2x) + 1.$

Step 2: Simplify the right-hand side:
$\log_3 7 \times \log_7 9$ can be written using the change of base formula:
$\log_3 7 = \frac{\log 7}{\log 3}$ and $\log_7 9 = \frac{\log 9}{\log 7}$. Multiplying these, we have:
$\frac{\log 9}{\log 3} = 2, \text{ since } \log 9 = \log 3^2 = 2 \log 3.$

Step 3: Equate and solve:
Equate the simplified versions:
$\log_5 (x^2 + 2x) + 1 = 2$
So, subtracting 1 from both sides:
$\log_5 (x^2 + 2x) = 1$
Taking antilogarithm, we find:
$x^2 + 2x = 5^1 = 5$

Rearrange to form a quadratic equation:
$x^2 + 2x - 5 = 0$

Step 4: Solve the quadratic equation:
Use the quadratic formula, where $a = 1$, $b = 2$, $c = -5$:
$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
$x = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot (-5)}}{2 \cdot 1} = \frac{-2 \pm \sqrt{4 + 20}}{2} = \frac{-2 \pm \sqrt{24}}{2} = \frac{-2 \pm 2\sqrt{6}}{2}$
$x = -1 \pm \sqrt{6}$

The valid answer must ensure $x + 2 > 0$, so $x = -1 + \sqrt{6}$.

Therefore, the solution to the problem is $x = -1 + \sqrt{6}$.

Let's solve the given equation step by step:

We start with:

$(2\log_3 2 + \log_3 x)\log_2 3 - \log_2 x = 3x - 7$

Firstly, use the change of base formula to convert $\log_2 3$ to base 3:

$\log_2 3 = \frac{\log_3 3}{\log_3 2} = \frac{1}{\log_3 2}$

Substitute this expression into the original equation:

$(2\log_3 2 + \log_3 x)\left(\frac{1}{\log_3 2}\right) - \log_2 x = 3x - 7$

Simplify the first term:

$\frac{2\log_3 2 + \log_3 x}{\log_3 2} = 2 + \frac{\log_3 x}{\log_3 2}$

Thus, the equation becomes:

$2 + \frac{\log_3 x}{\log_3 2} - \log_2 x = 3x - 7$

Convert $\log_2 x$ to base 3 using change of base:

$\log_2 x = \frac{\log_3 x}{\log_3 2}$

Substitute back into the equation:

$2 + \frac{\log_3 x}{\log_3 2} - \frac{\log_3 x}{\log_3 2} = 3x - 7$

The middle terms cancel out, simplifying to:

2 = 3x - 7

Solving for $x$:

Add 7 to both sides:

$9 = 3x$

Divide by 3:

$x = 3$

Thus, the solution to the problem is $x = 3$.