Multiplication of Logarithms: Using multiple rules

We will solve the problem step by step:

Step 1: Simplify $\log_9 e^3$

• Using the change of base formula, $\log_9 e^3 = \frac{\ln e^3}{\ln 9}$.
• We know $\ln e^3 = 3\ln e = 3$, because $\ln e = 1$.
• Thus, $\log_9 e^3 = \frac{3}{\ln 9} = \frac{3}{2\ln 3}$, since $\ln 9 = 2\ln 3$.
• Therefore, $\log_9 e^3 = \frac{3}{2\ln 3}$.

Step 2: Simplify $\log_2 24 - \log_2 8$

• Use the logarithm subtraction rule: $\log_2 24 - \log_2 8 = \log_2 \left(\frac{24}{8}\right) = \log_2 3$.

Step 3: Simplify $\ln 8 + \ln 2$

• Using the product property of logarithms: $\ln 8 + \ln 2 = \ln(8 \times 2) = \ln 16$.
• Since $16 = 2^4$, $\ln 16 = 4\ln 2$.

Step 4: Combine the results

• We need to check the overall structure: $\log_9 e^3 \times \log_2 3 \times 4 \ln 2$.
• Previously calculated: $\log_9 e^3 = \frac{3}{2 \ln 3}$, $\log_2 3 = \frac{\ln 3}{\ln 2}$.
• Therefore, the entire expression becomes:
• $\frac{3}{2 \ln 3} \times \frac{\ln 3}{\ln 2} \times 4 \ln 2 = \frac{3}{2} \times 4 = 6$.

Therefore, the solution to the problem is $6$.

To solve this problem, we'll carefully apply logarithmic properties:

• Step 1: Simplify the left-hand side:
  The left-hand side is given as $\log_64 \times \log_9x$. We simplify $\log_64$:
  $\log_64 = \frac{\log 4}{\log 6} = \frac{\log(2^2)}{\log 6} = \frac{2\log 2}{\log 6}$.
  Therefore, the left-hand side becomes $\frac{2\log 2}{\log 6} \times \log_9x$.
• Step 2: Simplify the right-hand side:
  The right-hand side is $(\log_6x^2 - \log_6x)(\log_92.5 + \log_91.6)$.
  First, simplify $\log_6x^2 - \log_6x = 2\log_6x - \log_6x = \log_6x$.
  For the other part, apply the product property: $\log_92.5 + \log_91.6 = \log_9(2.5 \times 1.6)$.
  Calculate $2.5 \times 1.6 = 4.0$, hence $\log_94$.
• Step 3: Equate and simplify:
  Now equate the simplified expressions: $\frac{2\log 2}{\log 6} \times \log_9x = \log_6x \cdot \log_94$.
  Change all logs to a common base (let's use natural log $\ln$) and solve:
• Step 4: Apply base conversion:
  $\log_9x = \frac{\ln x}{\ln 9}$, $\log_6x = \frac{\ln x}{\ln 6}$, and $\log_94 = \frac{\ln 4}{\ln 9}$.
• Step 5: Combine and solve:
  Perform algebraic manipulation and simplification:
  The equation becomes $\frac{2\ln 2}{\ln 6 \ln 9} \cdot \ln x = \frac{\ln x \cdot \ln 4}{\ln 6 \ln 9}$.
  Cancel $\ln x$ (non-zero due to $x > 0$) and solve for positive $x$.
• Conclude with the solution constraints:
  Given the properties and the domain involved, solution holds for all $0 < x$.

Therefore, the correct solution is: For all $0 < x$.

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 problem $\frac{2\log_7 8}{\log_7 4} + \frac{1}{\log_4 3} \times \log_2 9$, we will apply various logarithmic rules:

Step 1: Simplify $\frac{2\log_7 8}{\log_7 4}$.

• Using the power property, $\log_7 8 = \log_7 2^3 = 3\log_7 2$.
• Similarly, $\log_7 4 = \log_7 2^2 = 2\log_7 2$.
• The expression becomes $\frac{2 \times 3\log_7 2}{2\log_7 2} = 3$.

Step 2: Simplify $\frac{1}{\log_4 3} \times \log_2 9$.

• $\frac{1}{\log_4 3} = \log_3 4$, by inversion.
• $\log_2 9$ can be expressed as $\log_2 3^2 = 2\log_2 3$.
• The product becomes $\log_3 4 \times 2\log_2 3 = 2 \cdot \frac{\log_2 4}{\log_2 3} \times \log_2 3$.
• Since $\log_2 4 = 2$, this simplifies to $2 \times \frac{2}{1} = 4$.

Step 3: Add the results from Steps 1 and 2:
$3 + 4 = 7$.

Therefore, the solution to the problem is $7$.

To solve this problem, we'll proceed as follows:

• Step 1: Rewrite each logarithmic expression using the change of base formula.
• Step 2: Simplify the expressions using properties of logarithms.
• Step 3: Identify the final expression.

Now, let's work through each step:

Step 1: We begin by converting each logarithm to the natural logarithm base.
Using the change of base formula, we have:

$\frac{\log_3 11}{\log_3 4} = \frac{\frac{\ln 11}{\ln 3}}{\frac{\ln 4}{\ln 3}} = \frac{\ln 11}{\ln 4}$.

Step 2: Next, simplify the second expression:

$\frac{1}{\ln 3} \cdot 2\log 3 = 2$.

This follows because $\log 3$ in natural logarithms converts to $\ln 3$, and thus:

$\frac{2\ln 3}{\ln 3} = 2$.

Hence, our entire expression now is $\frac{\ln 11}{\ln 4} + 2$.

Step 3: Express $2$ as a logarithm. Using the properties of logarithms:

$2 = \log e^2$, since $\ln e = 1$.

Therefore, the entire expression becomes:

$\frac{\ln 11}{\ln 4} + \log e^2$.

By the properties of logarithms, this can also be expressed as:

$\log_4 11 + \log e^2$.

Thus, the expression simplifies directly to:

$\log_4 11 + \log e^2$.

Therefore, the solution to the problem is $\log_4 11 + \log e^2$.

To solve this problem, we'll simplify the expression step-by-step, using algebraic rules for logarithms:

• Step 1: Simplify the numerator $\frac{\log_7 6 - \log_7 1.5}{3 \log_7 2}$

First, apply the logarithm quotient rule to the numerator:
$\log_7 6 - \log_7 1.5 = \log_7 \left(\frac{6}{1.5}\right) = \log_7 4$

• Step 2: Simplify $3 \log_7 2$ in the denominator.

The denominator is $3 \times \log_7 2$.

• Step 3: Address the next part of the expression: $\frac{1}{\log_{\sqrt{8}} 2}$.

By changing the base, use $\log_{\sqrt{8}} 2 = \frac{\log_{8} 2}{\frac{1}{2}}$ because $\sqrt{8} = 8^{1/2}$. Now, $\log_8 2 = \frac{1}{3}$ as $8^{1/3} = 2$. So, $\log_{\sqrt{8}} 2 = \frac{\log_2 8}{1/2} = \frac{1/3}{1/2} = \frac{2}{3}$.

Therefore, the reciprocal is $\frac{1}{\log_{\sqrt{8}} 2} = \frac{3}{2}$.

• Step 4: Combine and simplify the expression.

The complete logarithmic expression simplifies as follows:
$\frac{\log_7 4}{3 \log_7 2} \cdot \frac{3}{2} = \frac{\log_7 (2^2)}{3 \log_7 2} \cdot \frac{3}{2}$

Using the power rule, $\log_7 4 = 2 \log_7 2$. Plug this back into the expression:
$\frac{2 \log_7 2}{3 \log_7 2} \cdot \frac{3}{2}$
The $\log_7 2$ cancels within the fraction, and we are left with $\frac{2}{3} \times \frac{3}{2} = 1$.

Therefore, the solution to the problem is $1$.

To solve the problem, we must evaluate the expression $\frac{1}{\ln 4} \cdot \frac{1}{\log_8 10}$.

First, convert $\log_8 10$ using the change of base formula. We have:

• $\log_8 10 = \frac{\ln 10}{\ln 8}$.

Substitute this back into the original expression:

$\frac{1}{\ln 4} \cdot \frac{1}{\log_8 10} = \frac{1}{\ln 4} \cdot \frac{\ln 8}{\ln 10}$.

Next, we need to simplify the expression. We know that $\ln 8 = \ln (2^3) = 3 \ln 2$ and $\ln 4 = \ln (2^2) = 2 \ln 2$.

Substitute these into the expression:

= $\frac{1}{2 \ln 2} \cdot \frac{3 \ln 2}{\ln 10}$.

Simplify by canceling $\ln 2$:

= $\frac{3}{2} \cdot \frac{1}{\ln 10}$.

Now express $\ln 10 = \ln (e \cdot \log e)$, meaning this is equivalent to $\log e$. Continuing, the expression $\frac{3}{2} \cdot \frac{1}{\log e} = \frac{3}{2} \log e$.

Therefore, the simplified solution to the given expression is $\frac{3}{2} \log e$.

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

• Step 1: Convert the logarithms into another base using the change of base rule.
• Step 2: Simplify $\ln e$ since $\ln e = 1$.
• Step 3: Simplify the expression using known values.
• Step 4: Solve the equation for $x$.

Now, let's work through each step:

Step 1: Given the equation $\log_3 x^2 \log_5 27 - \log_5 8 = \ln e$, we know that $\ln e = 1$. We will first simplify the right side to get:
$\log_3 x^2 \log_5 27 - \log_5 8 = 1$

Step 2: Use the change of base formula.

Using $\log_b a = \frac{\ln a}{\ln b}$, rewrite $\log_5 27$ and $\log_5 8$:

$\log_5 27 = \frac{\ln 27}{\ln 5} \quad \text{and} \quad \log_5 8 = \frac{\ln 8}{\ln 5}$

Plug in the values:

$\log_3 x^2 \frac{\ln 27}{\ln 5} - \frac{\ln 8}{\ln 5} = 1$

Step 3: Multiply through by $\ln 5$ to eliminate the denominators:
$\log_3 x^2 \ln 27 - \ln 8 = \ln 5$

Now knowing $\ln 27 = 3\ln 3$, solve the equation:

$\log_3 x^2 = \frac{\ln 5 + \ln 8}{3 \ln 3}$

Apply the logarithm base rule:

$x^2 = 3^{\left(\frac{\ln 5 + \ln 8}{3\ln 3}\right)}$

Step 4: Simplify and solve for $x$. Recognize this exponent could become $\frac{\ln 40}{3\ln 3}$:

$x^2 = 3^{\frac{\ln 40}{3\ln 3}} = 40^{1/3}$

Finally, solve for $x$:

$x = \pm \sqrt[6]{40}$

Therefore, the solution to the problem is $x = \pm\sqrt[6]{40}$.

Let's solve the problem step-by-step:

We start with the equation:

We simplify the right side using the product rule for logarithms:

Next, we simplify $\log_5 8$ on the left side:

Thus, we substitute into the original equation:

Now, divide both sides by $3 \log_5 2$:

Using the change of base formula, express $\log_5 (2a^2)$ and $\log_5 2$ with base 2:

Substitute these into the equation:

This implies:

Raising 2 to both sides of the equation to remove the logarithms:

Therefore, solving for $x$:

Thus, we conclude:

Therefore, the value of $x$ in terms of $a$ is $\sqrt[3]{\frac{2a^2}{27}}$.

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}.

To solve the given problem, we begin by simplifying each component of the expression.

Step 1: Simplify $\frac{\log_8x^3}{\log_8x^{1.5}}$.
Applying the power rule of logarithms, we get:
$\log_8x^3 = 3 \log_8x$, and $\log_8x^{1.5} = 1.5 \log_8x$.
Thus, $\frac{3 \log_8x}{1.5 \log_8x} = \frac{3}{1.5} = 2$.

Step 2: Simplify $\frac{1}{\log_{49}x} \times \log_7x^5$.
First, notice that $\log_7x^5 = 5 \log_7x$ by the power rule.
Applying the change of base formula, $\log_{49}x = \frac{\log_7x}{\log_749} = \frac{\log_7x}{2}$ because $49 = 7^2$.
This gives $\frac{1}{\log_{49}x} = \frac{2}{\log_7x}$.
Therefore, $\frac{2}{\log_7x} \times 5 \log_7x = 2 \times 5 = 10$.

Step 3: Combine the results from Step 1 and Step 2.
The simplified expression is $2 + 10 = 12$.

Therefore, the solution to the problem is $12$.

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

• Step 1: Express $\log_4{7}$ and $\log_{\frac{1}{49}}{a}$ using the change-of-base formula.
• Step 2: Simplify the product $\log_4{7} \times \log_{\frac{1}{49}}{a}$.
• Step 3: Simplify the entire expression by using logarithmic identities.

Let's work through each step:
Step 1: Using the change-of-base formula, $\log_4{7} = \frac{\log_k{7}}{\log_k{4}}$ and $\log_{\frac{1}{49}}{a} = \frac{\log_k{a}}{\log_k{\frac{1}{49}}}$. Choose $k = 10$ (common log) for simplicity.
Note that $\log_k{\frac{1}{49}} = \log_k{49^{-1}} = -\log_k{49}$. Also, $49 = 7^2$, so $\log_k{49} = 2\log_k{7}$. Therefore, $\log_{\frac{1}{49}}{a} = \frac{\log_k{a}}{-2\log_k{7}}$.

Step 2: The product $\log_4{7} \times \log_{\frac{1}{49}}{a} = \left(\frac{\log_k{7}}{\log_k{4}}\right)\left(\frac{\log_k{a}}{-2\log_k{7}}\right)$ simplifies to $\frac{\log_k{a}}{-2\log_k{4}}$ after canceling $\log_k{7}$.

Step 3: The expression becomes $\frac{\frac{\log_k{a}}{-2\log_k{4}}}{c\log_4{b}}$, which simplifies to $\frac{\log_k{a}}{-2c\log_k{4}\log_4{b}}$. Convert $\log_4{b}$ into $\frac{\log_k{b}}{\log_k{4}}$, leading to $\frac{\log_k{a}}{-2c\log_k{b}}$. Using the change-of-base formula again, this gives $-\frac{1}{2}\log_{b^c}{a}$.

This can be rewritten using inverse log properties as $\log_{b^c}{\left(\frac{1}{\sqrt{a}}\right)}$.

Therefore, the solution to the problem is $\log_{b^c}\frac{1}{\sqrt{a}}$.

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 $x = 3$.

To solve the given equation, we need to simplify the logarithmic expressions and then solve for $x$. Let's proceed with the given equation:

$\frac{1}{\log_x 3} \times x^2 \log_{1/x} 27 + 4x + 6 = 0$

Step 1: Simplify the logarithmic terms.

Apply the change of base formula to the logarithms:

$\log_x 3 = \frac{\ln 3}{\ln x}$

Thus, $\frac{1}{\log_x 3} = \frac{\ln x}{\ln 3}$.

For the second logarithmic term: $\log_{1/x} 27 = -\log_x 27 = -\frac{\ln 27}{\ln x}$.

Step 2: Substitute these simplifications back into the equation.

We have:

$\frac{\ln x}{\ln 3} \times x^2 \times -\frac{\ln 27}{\ln x} + 4x + 6 = 0$

Simplify this expression:

The $\ln x$ terms cancel each other out in the expression $\frac{\ln x}{\ln 3} \times x^2 \times -\frac{\ln 27}{\ln x}$.

Thus, it becomes:

$-\frac{\ln 27}{\ln 3} x^2 + 4x + 6 = 0$

The value of $-\frac{\ln 27}{\ln 3}$ is actually $-\log_3 27 = -3$ because $27 = 3^3$.

Therefore, the simplified equation is:

$-3x^2 + 4x + 6 = 0$

Step 3: Solve the quadratic equation.

Rearrange it to $3x^2 - 4x - 6 = 0$.

Apply the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

Here, $a = 3$, $b = -4$, $c = -6$.

So, the solution becomes:

$x = \frac{4 \pm \sqrt{(-4)^2 - 4 \times 3 \times (-6)}}{2 \times 3}$

This simplifies to:

$x = \frac{4 \pm \sqrt{16 + 72}}{6}$

$x = \frac{4 \pm \sqrt{88}}{6}$

Simplify $\sqrt{88} = \sqrt{4 \times 22} = 2\sqrt{22}$.

Thus,

$x = \frac{4 \pm 2\sqrt{22}}{6}$

Simplifying further gives us:

$x = \frac{2 \pm \sqrt{22}}{3}$

The valid positive solution (since logarithms are not satisfied with negative bases) is: