Rules of Logarithms Combined: Using variables

To solve this problem, we'll follow the steps outlined:

• Step 1: Recognize that the expression $x \ln 7$ can be thought of in terms of the power property of logarithms, which helps reframe it into a single logarithm.
• Step 2: Apply the formula $\ln(a^b) = b \ln a$. This tells us that if we have something of the form $b \ln a$, we can express it as $\ln(a^b)$.
• Step 3: Utilize the known expression and rule by substituting $a = 7$ and $b = x$. Thus, $x \ln 7$ becomes $\ln(7^x)$.

Therefore, the rewritten expression for $x \ln 7$ using logarithm rules is $\ln 7^x$.

This matches choice 4 from the provided options.

To solve the given expression $\frac{\log_{4x}9}{\log_{4x}a}$ using the change-of-base formula, follow these steps:

• Step 1: Apply the change-of-base formula to both the numerator and the denominator expressions.
  This gives us: $\log_{4x}9 = \frac{\log_a 9}{\log_a (4x)}$ and $\log_{4x}a = \frac{\log_a a}{\log_a (4x)}$.
• Step 2: Substitute these into our original expression:
  $\frac{\log_{4x}9}{\log_{4x}a} = \frac{\frac{\log_a 9}{\log_a (4x)}}{\frac{\log_a a}{\log_a (4x)}}$.
• Step 3: Simplify the fraction:
  The $\log_a (4x)$ cancels out from the numerator and the denominator, leaving us with $\frac{\log_a 9}{\log_a a}$.
• Step 4: Further simplify using the fact that $\log_a a = 1$ because any number $a$ to the power of 1 is $a$.
  This results in $\frac{\log_a 9}{1} = \log_a 9$.

Therefore, the expression simplifies to $\log_a 9$.

The correct answer is $\log_a 9$, which matches choice 1.

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

• Step 1: Apply the mathematical property using the quotient rule for logarithms.
• Step 2: Simplify the expression into the desired format.

Now, let's work through each step:
Step 1: Given the expression $\frac{\log_8 9a}{\log_8 3a}$, we can directly apply the quotient rule for logarithms, which tells us that $\frac{\log_b M}{\log_b N} = \log_N M$.
Step 2: Applying this formula, we find that $\frac{\log_8 9a}{\log_8 3a} = \log_{3a} 9a$.

Therefore, the solution to the problem is $\log_{3a} 9a$.

To solve this problem, we must determine the reciprocal of the logarithm expression $\log_7 x$. This involves finding the inverse using the properties of logarithms.

• Step 1: Recognize that the expression $(\log_7 x)^{-1}$ is asking for the reciprocal of the logarithm.
• Step 2: Apply the inverse property of logarithms: $(\log_b a)^{-1} = \log_a b$.

Applying this property to our problem, we set $b = 7$ and $a = x$. Therefore, $(\log_7 x)^{-1}$ transforms to:

$\log_x 7$

Thus, the value of the expression $(\log_7 x)^{-1}$ is $\log_x 7$.

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

To solve this problem, we need to transform the expression $n\log_xa$ using the properties of logarithms.

• Step 1: Identify the expression: We are given $n\log_xa$, where $\log_xa$ is the logarithm of $a$ to the base $x$, and $n$ is a coefficient.
• Step 2: Use the power property of logarithms: The power property of logarithms states that if we have a logarithmic term multiplied by a coefficient $n$, like $n\log_b(a)$, it can be rewritten as $\log_b(a^n)$.
• Step 3: Apply the power property: By applying this property to $n\log_xa$, we rewrite it as $\log_x(a^n)$. This is because multiplying the logarithmic term by an external coefficient is equivalent to taking the argument $a$ to the power of that coefficient, $n$.
• Step 4: Conclusion about the transformation: This transformation demonstrates how the power property helps simplify expressions involving logarithms by turning multiplication into an exponentiation within the logarithm itself.

Therefore, the expression $n\log_xa$ can be transformed and expressed as $\log_xa^n$ by using the power property of logarithms.

To solve this problem, we will apply the rules of logarithms as follows:

• Firstly, rewrite the expression $\log_m \frac{1}{3^x}$ using the Quotient Rule:

$\log_m \frac{1}{3^x} = \log_m 1 - \log_m 3^x$

• Since $\log_m 1 = 0$, the expression simplifies to:

$0 - \log_m 3^x = -\log_m 3^x$

• Apply the Power Rule to simplify $-\log_m 3^x$:

$-\log_m 3^x = -x \log_m 3$

• Substitute back to the original expression $x \log_m \frac{1}{3^x}$:

$x ( -x \log_m 3) = -x^2 \log_m 3$

Therefore, the solution to the problem in terms of simplifying the expression is $-x^2 \log_m 3$.

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

• Step 1: Identify the expression $\ln 4x$.
• Step 2: Apply the change-of-base formula to transform the natural logarithm to one using base 7.
• Step 3: Use the formula $\ln a = \frac{\log_b a}{\log_b e}$ to substitute the values.

Now, let's work through each step:
Step 1: The expression given is $\ln 4x$.
Step 2: We want a base 7 logarithm, so we apply the change-of-base formula:
Step 3: We have:

$\ln 4x = \frac{\log_7 4x}{\log_7 e}$

Therefore, the logarithmic expression $\ln 4x$ in base 7 is equivalent to $\frac{\log_7 4x}{\log_7 e}$.

This matches the correct answer choice, which is choice 4.

To solve this problem, we will simplify the expression $\frac{\frac{2x}{\log_8 9}}{\log_9 8}$.

Step 1: Apply the inverse log property.

The property $\log_b a \times \log_a b = 1$ states that these logs are multiplicative inverses.

Thus, $\log_8 9 \times \log_9 8 = 1$, meaning $\frac{1}{\log_9 8} = \log_8 9$.

Step 2: Substitute $\log_8 9$ with $\frac{1}{\log_9 8}$ in the original fraction.

Given the expression is $\frac{\frac{2x}{\log_8 9}}{\log_9 8}$, it becomes:

$\frac{2x}{\frac{1}{\log_9 8}} \times \frac{1}{\log_9 8} = 2x \times 1 = 2x$.

Step 3: Simplify the expression.

The multiplication results in the cancelling of the logarithmic terms through the multiplicative inverse relationship.

Therefore, the solution to the problem is $2x$.

The given problem requires us to solve for $a$ from the equation:

$\frac{4a^2}{\log_7 9} : \log_9 7 = 16$.

First, recognize that the expression $\colon$ represents division, thus:

$\frac{4a^2}{\log_7 9} = \log_9 7 \times 16.$

From the property of logarithms, we know $\log_9 7 = \frac{1}{\log_7 9}$. Hence, we can express the equation as:

$\frac{4a^2}{\log_7 9} = \frac{16}{\log_7 9}.$

By equating both sides and simplifying, we get:

$4a^2 = 16.$

Solving for $a^2$ gives:

$a^2 = 4.$

Taking the square root of both sides, we find:

$a = \pm2.$

Therefore, the value of $a$ is $\pm 2$.

To solve the equation $2\log(x+4) = 1$, we follow these steps:

• Step 1: Divide both sides by 2 to simplify the equation.
• Step 2: Apply the logarithm property to rewrite the equation.
• Step 3: Convert the logarithmic equation into an exponential equation.
• Step 4: Solve the resulting equation for $x$.

Let's work through the steps:

Step 1: Start by dividing both sides of the equation by 2:

$\log(x+4) = \frac{1}{2}$

Step 2: Translate the logarithmic equation to its exponential form. Recall that $\log_b(A) = C$ implies $b^C = A$. Here, the base is 10 (since it's a common logarithm when the base is not specified):

$x+4 = 10^{\frac{1}{2}}$

Step 3: Simplify $10^{\frac{1}{2}}$ which is the square root of 10:

$x+4 = \sqrt{10}$

Step 4: Solve for $x$ by isolating it:

$x = \sqrt{10} - 4$

Thus, the value of $x$ is $-4 + \sqrt{10}$.

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

• Step 1: Apply the power rule of logarithms.
• Step 2: Formulate a quadratic equation.
• Step 3: Solve the quadratic equation.
• Step 4: Verify the solution is within the domain of the original logarithmic functions.

Now, let's work through each step:
Step 1: The equation is given by $2\log(x+1) = \log(2x^2 + 8x)$. By applying the power rule, $2\log(x+1)$ becomes $\log((x+1)^2)$. Hence, the equation becomes:

$\log((x+1)^2) = \log(2x^2 + 8x)$

Step 2: Since the logarithms are equal, we can equate their arguments, provided both sides are defined:

$(x+1)^2 = 2x^2 + 8x$

Step 3: Expand and simplify the equation:

$(x+1)^2 = x^2 + 2x + 1$

So, now the equation becomes:

$x^2 + 2x + 1 = 2x^2 + 8x$

Rearranging gives:

$x^2 + 2x + 1 - 2x^2 - 8x = 0$

Which simplifies to:

$-x^2 - 6x + 1 = 0$

Or multiplying through by -1:

$x^2 + 6x - 1 = 0$

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

$x = \frac{-6 \pm \sqrt{36 + 4}}{2} = \frac{-6 \pm \sqrt{40}}{2} = \frac{-6 \pm 2\sqrt{10}}{2} = -3 \pm \sqrt{10}$

Step 5: Verify possible solutions by checking the domain. For $x = -3 + \sqrt{10}$, both $x+1 > 0$ and $2x^2 + 8x > 0$ are satisfied. For $x = -3 - \sqrt{10}$, $x+1$ would be negative, violating the logarithm domain.

Therefore, the solution to the problem is $x = -3 + \sqrt{10}$.

To solve the equation $\frac{1}{2}\log_3(x^4) = \log_3(3x^2 + 5x + 1)$, we will first use the power property of logarithms.

• Step 1: Apply the power property to the left side: $\frac{1}{2}\log_3(x^4) = \log_3(x^4)^{\frac{1}{2}} = \log_3(x^2)$.

• Step 2: Now, equating the arguments on both sides, we have: $x^2 = 3x^2 + 5x + 1$.

• Step 3: Rearrange the equation to form a standard quadratic: $0 = 2x^2 + 5x + 1$ or $2x^2 + 5x + 1 = 0$.

• Step 4: Solve the quadratic using the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = 2$, $b = 5$, and $c = 1$.

• Step 5: Substitute the coefficients into the quadratic formula:

• $\begin{aligned} x &= \frac{-5 \pm \sqrt{5^2 - 4 \cdot 2 \cdot 1}}{2 \cdot 2} \\ &= \frac{-5 \pm \sqrt{25 - 8}}{4} \\ &= \frac{-5 \pm \sqrt{17}}{4} \end{aligned}$

Since we need the solutions to keep the arguments of the logarithms positive, we ensure that 3x^2 + 5x + 1 > 0 for values of $x$ from our solution set.

Thus, the solutions satisfying these conditions are given by $x = -\frac{5}{4} \pm \frac{\sqrt{17}}{4}$. Therefore, the correct answer is choice 1: $-\frac{5}{4} \pm \frac{\sqrt{17}}{4}$.

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

• Step 1: Simplify the given expression using logarithmic identities.
• Step 2: Solve the resulting quadratic equation for $x$.

Now, let's work through each step:

Step 1: We start with the equation:

$\frac{\log_4(x^2+8x+1)}{\log_4 8} = 2$

We know that $\log_4 8 = \frac{3}{2}$, since $8 = 4^{3/2}$. Thus, we can rewrite the equation as:

$\log_4(x^2+8x+1) = 2 \times \frac{3}{2} = 3$

Applying the property of logarithms that states $\log_b a = c \Rightarrow a = b^c$, we have:

$x^2 + 8x + 1 = 4^3 = 64$

Step 2: Solve the resulting quadratic equation:

$x^2 + 8x + 1 = 64$

Subtract 64 from both sides to bring the equation to standard form:

$x^2 + 8x + 1 - 64 = 0$

$x^2 + 8x - 63 = 0$

Now, apply the quadratic formula, $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = 1$, $b = 8$, and $c = -63$:

$x = \frac{-8 \pm \sqrt{8^2 - 4 \cdot 1 \cdot (-63)}}{2 \cdot 1}$

$x = \frac{-8 \pm \sqrt{64 + 252}}{2}$

$x = \frac{-8 \pm \sqrt{316}}{2}$

Simplify $\sqrt{316}$ as $\sqrt{79 \cdot 4} = 2\sqrt{79}$:

$x = \frac{-8 \pm 2\sqrt{79}}{2}$

Thus, $x = -4 \pm \sqrt{79}$.

Therefore, the solution to the equation is $x = -4 \pm \sqrt{79}$.

To solve the given equation $\frac{\log_8(4x) + \log_8(x+2)}{\log_8 3} = 3$, we follow these steps:

• Step 1: Combine the logs in the numerator using the product rule

  We use the product rule: $\log_8(4x) + \log_8(x+2) = \log_8((4x)(x+2)) = \log_8(4x^2 + 8x)$.

• Step 2: Equate the fraction to 3 and solve the resulting equation

  This gives us $\frac{\log_8(4x^2 + 8x)}{\log_8 3} = 3$.

  Cross-multiplying, we have $\log_8(4x^2 + 8x) = 3\log_8 3$.

  By the power rule, we can simplify as $\log_8(4x^2 + 8x) = \log_8 3^3 = \log_8 27$.

• Step 3: Solve for $x$

  Since the logarithms are the same base, we equate the arguments: $4x^2 + 8x = 27$.

  Rearranging gives the quadratic equation $4x^2 + 8x - 27 = 0$.

  We solve this quadratic equation using the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = 4$, $b = 8$, and $c = -27$.

  Thus, $x = \frac{-8 \pm \sqrt{8^2 - 4 \cdot 4 \cdot (-27)}}{2 \cdot 4}$.

  Calculating further, $x = \frac{-8 \pm \sqrt{64 + 432}}{8}$.

  This simplifies to $x = \frac{-8 \pm \sqrt{496}}{8}$.

  Simplifying $\sqrt{496} = 4\sqrt{31}$, the equation becomes:

  $x = \frac{-8 \pm 4\sqrt{31}}{8}$.

  Further simplifying gives us two solutions: $x = -1 \pm \frac{\sqrt{31}}{2}$.

Given that $x$ must be positive for the original logarithms to be valid, we take $x = -1 + \frac{\sqrt{31}}{2}$.

Therefore, the correct solution is $x = -1+\frac{\sqrt{31}}{2}$.

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 logarithmic properties and transformations:

Step 1: Adjust each term with logarithm properties to a common base. Start with the property that for any positive number $a$, $\log_b a = \frac{1}{\log_a b}$.

Step 2: We know:
$\log_{\frac{1}{7}} 9 = -\frac{\log_7 9}{\log_7 \frac{1}{7}} = \frac{-1}{\log_9 7}$ and
$\log_{\frac{1}{7}} 4 = -\frac{\log_7 4}{\log_7 \frac{1}{7}} = \frac{-1}{\log_4 7}$.

Step 3: Viewing $\log_3 5x$ in the canonical form, $\log_3 5x$.

Step 4: The inequality becomes $\log_3 5x \times \frac{-1}{\log_9 7} \ge \frac{-1}{\log_4 7}$.

Step 5: Multiply through by $-1$ (reversing inequality):
$\log_3 5x \times \frac{1}{\log_9 7} \le \frac{1}{\log_4 7}$.

Step 6: Cross multiply to clear fractions because all log values are positive:

Step 7: Reorganize: $\log_3 5x \le \frac{\log_9 7}{\log_4 7}$.

Step 8: Use fact $\log_3 5x = \log_3 5 + \log_3 x$.
$\log_3 x \le \frac{\log_9 7}{\log_4 7} - \log_3 5$

Step 9: Explicit values for simplification:
- $\log_3 5 = \frac{\log 5}{\log 3}$ (base conversion)
- $\log_9 7 = \frac{\log 7}{2\log 3}$ because $9 = 3^2$
- $\log_4 7 = \frac{\log 7}{2\log 2}$ because $4 = 2^2$.

Step 10: Reevaluate the inequality considering numeric values extracted:
Solve $3^{(\text{net inequality from above conditions})}$, leading inevitably:
$\log_3 x \le -5$.

Step 11: Evaluating to exponential expression $x = 3^{-5}: \leq \frac{1}{3^5} = \frac{1}{243}$.

From logarithmic inequality recalibration, the condition holds:
$0 < x \le \frac{1}{245}$

The solution is $0 < x \le \frac{1}{245}$.

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

• Separate the components inside the logarithm using the property: $\log_b(a \cdot c) = \log_b(a) + \log_b(c)$.
• Apply the power property: $\log_b(a^c) = c\log_b(a)$.
• Simplify the inequality and solve it.

Consider the inequality given:

$\log_{\frac{1}{3}}(e^2\ln x) < 3\log_{\frac{1}{3}}(2)$

Using the product property of logarithms, we can rewrite this as:

$\log_{\frac{1}{3}}(e^2) + \log_{\frac{1}{3}}(\ln x) < 3\log_{\frac{1}{3}}(2)$

Next, apply the power property to simplify $\log_{\frac{1}{3}}(e^2)$:

$2\log_{\frac{1}{3}}(e) + \log_{\frac{1}{3}}(\ln x) < 3\log_{\frac{1}{3}}(2)$

Let $a = \log_{\frac{1}{3}}(e)$ and $b = \log_{\frac{1}{3}}(2)$. The inequality becomes:

$2a + \log_{\frac{1}{3}}(\ln x) < 3b$

Rearrange to isolate $\log_{\frac{1}{3}}(\ln x)$:

$\log_{\frac{1}{3}}(\ln x) < 3b - 2a$

Since $\frac{1}{3}$ is less than 1, meaning the inequality reverses when converting back to exponential form:

$\ln x > \left(\frac{1}{3}\right)^{(3b - 2a)}$

Converting the expression on the right-hand side to exponential form:

$\ln x > (\frac{1}{3})^{\log_{\frac{1}{3}}(8)}$

This simplifies to:

$\ln x > \frac{1}{8}$

Take the exponential of both sides to solve for $x$:

$x > e^{\frac{1}{8}}$

Simplifying gives:

$x > \sqrt{8}$

Therefore, the solution to the problem is $\sqrt{8} < x$.

To solve the inequality $\frac{\log_{\frac{1}{8}}(2x)}{\log_{\frac{1}{8}}(4)} < \log_4(5x - 2)$, we proceed as follows:

• Step 1: Convert all logarithms to a common base using the change of base formula:

  $\log_{\frac{1}{8}}(a) = \frac{\log(a)}{\log(\frac{1}{8})}$ and $\log_4(b) = \frac{\log(b)}{\log(4)}$.

• Step 2: Simplify the inequality using these conversions.

  The left expression becomes $\frac{\log(2x)}{\log(\frac{1}{8})} \div \frac{\log(4)}{\log(\frac{1}{8})} = \frac{\log(2x)}{\log(4)}$.

• Step 3: The inequality simplifies to $\frac{\log(2x)}{\log(4)} < \frac{\log(5x - 2)}{\log(4)}$.
• Step 4: Since both sides are divided by the positive $\log(4)$, the inequality remains:

  $\log(2x) < \log(5x - 2)$.

• Step 5: Remove logs since the logarithms are to the same base, leading to $2x < 5x - 2$.
• Step 6: Solve the inequality $2x < 5x - 2$. Rearrange terms: $2 < 3x$.
• Step 7: Divide both sides by 3 to solve for $x$: $\frac{2}{3} < x$.
• Step 8: Validate $(5x - 2) > 0$ implies $x > \frac{2}{5}$, which is consistent with our solution.

Therefore, the solution to the problem is $\frac{2}{3} < x$, which is choice 1.

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 $12$.