Rules of Logarithms Combined: Using multiple rules

To solve the equation $\log 4x + \log 2 - \log 9 = \log_2 4$, we will follow these steps:

• Step 1: Simplify the left side using logarithmic properties
• Step 2: Convert the right side using change of base
• Step 3: Equate the simplified expressions and solve for $x$

Step 1: Simplify the left side:

The left side $\log 4x + \log 2 - \log 9$ can be combined using the properties of logarithms:

$\log 4x + \log 2 = \log(4x \cdot 2) = \log(8x)$

Now, using the subtraction property:

$\log (8x) - \log 9 = \log \left(\frac{8x}{9}\right)$

Step 2: Convert the right side using the change of base formula:

$\log_2 4 = \frac{\log 4}{\log 2}$

We recognize that $4 = 2^2$, so $\log_2 4 = 2$.

Step 3: Equate the expressions and solve for $x$:

Now equate:

$\log \left(\frac{8x}{9}\right) = 2$

This implies:

$\frac{8x}{9} = 10^2 = 100$

Thus, solving for $x$:

$8x = 900$

$x = \frac{900}{8} = 112.5$

Therefore, the solution to the problem is $x = 112.5$.

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 follow these steps:

• Step 1: Combine the logarithms in the numerator.
• Step 2: Simplify the expression using logarithmic properties.

Now, let's work through each step:

Step 1: Combine the logarithms in the numerator using the sum of logarithms property:

$\log_45 + \log_42 = \log_4(5 \times 2) = \log_4 10.$

Step 2: Simplify the entire expression $\frac{\log_4 10}{3\log_4 2}$:

$\frac{\log_4 10}{3 \log_4 2} = \frac{\log_4 10}{\log_4 2^3} = \frac{\log_4 10}{\log_4 8} = \log_8 10.$

This follows from the property that $\frac{\log_b x}{\log_b y} = \log_y x$.

Therefore, the solution to the problem is $\log_8 10$.

Defined domain

x>0

x+1>0

x>-1

$\log7x+\log\left(x+1\right)-\log7=\log2x-\log x$

$\log\frac{7x\cdot\left(x+1\right)}{7}=\log\frac{2x}{x}$

We reduce by: $7$ and by $X$

$x\left(x+1\right)=2$

$x^2+x-2=0$

$\left(x+2\right)\left(x-1\right)=0$

$x+2=0$

$x=-2$

Undefined domain x>0

$x-1=0$

$x=1$

Defined domain

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 this problem, we'll follow these steps:

• Step 1: Apply the change-of-base formula to $\frac{\ln 4}{\ln 5}$.

• Step 2: Apply the reciprocal property to $\frac{1}{\log_6 5}$.

• Step 3: Use the subtraction property of logs to simplify the expression.

• Step 4: Combine the simplified logarithms and multiply by -3.

Now, let's work through each step:

Step 1: Using the change-of-base formula, we have $\frac{\ln 4}{\ln 5} = \log_5 4$.

Step 2: Apply the reciprocal property to the third term: $\frac{1}{\log_6 5} = \log_5 6$.

Step 3: Substitute into the expression: $-3(\log_5 4 - \log_5 7 + \log_5 6)$.

Step 4: Combine terms using the properties of logs: $\log_5 4 - \log_5 7 + \log_5 6 = \log_5 \left(\frac{4 \times 6}{7}\right)$.

Step 5: Simplify to get: $\log_5 \left(\frac{24}{7}\right)$.

Multiply by -3: $-3(\log_5 (\frac{24}{7})) = 3\log_5 \left(\frac{7}{24}\right)$.

Therefore, the solution to the problem is $3\log_5 \frac{7}{24}$.

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}.

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 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 $-\frac{1}{2}\log_{b^c}{a}$.