Power Property of Logorithms: Applying the formula

To solve this problem, let's simplify $2\log_3 8$ using logarithm rules.

• Step 1: Recognize the expression form
  The expression is of the form $a \cdot \log_b c$, where $a = 2$, $b = 3$, and $c = 8$.
• Step 2: Apply the power property
  According to the power property of logarithms, $2 \cdot \log_3 8$ can be simplified to $\log_3 (8^2)$.
• Perform the calculation
  Calculate $8^2$, which is $64$.
• Step 3: Simplify further
  Therefore, we have $\log_3 64$.

This is a straightforward application of the power property of logarithms. By applying this property correctly, we've simplified the original expression correctly.

Therefore, the simplified form of $2\log_3 8$ is $\log_3 64$.

To simplify the expression $3\log_76$, we apply the power property of logarithms, which states:

$a\log_b c = \log_b(c^a)$

Step 1: Identify the given expression: $3\log_76$.

Step 2: Apply the power property of logarithms:

$3\log_76 = \log_7(6^3)$

Step 3: Calculate $6^3$:

$6^3 = 6 \times 6 \times 6 = 36 \times 6 = 216$

Step 4: Substitute back into the logarithmic expression:

$\log_7(6^3) = \log_7216$

Therefore, the simplified expression is $\log_7216$.

Comparing with the answer choices, the correct choice is:

$\log_7216$

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 problem $\log_6 8$, we need to express the number 8 as a power of a base that simplifies the logarithm. We can write 8 as $2^3$, because 8 equals 2 multiplied by itself three times.

Let's use the power property of logarithms, which is:

• $\log_b (a^n) = n \log_b a$

Applying this property to $\log_6 8$, we have:

$\log_6 8 = \log_6 (2^3)$

Using the power property, this becomes:

$\log_6 (2^3) = 3 \log_6 2$

Therefore, the expression for $\log_6 8$ in terms of $\log_6 2$ is:

$3 \log_6 2$.

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 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 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 this problem, we need to compare the expressions $\log_4 x \times \log_5 64$ and $\log_5 (x^3 + x^2 + x + 1)$.

First, calculate $\log_5 64$. We know that $64 = 4^3 = 2^6$. Therefore:
$\log_5 64 = \frac{\log_5 2^6}{\log_5 4} = \frac{6 \log_5 2}{2 \log_5 2} = 3$

Next, simplify the left-hand side expression $\log_4 x$. Using the change of base formula:
$\log_4 x = \frac{\log_5 x}{\log_5 4}$

Therefore, the left-hand side becomes:
$\frac{\log_5 x}{\log_5 4} \times 3 = \frac{3 \log_5 x}{2 \log_5 2}$

For the inequality:
$\frac{3 \log_5 x}{2 \log_5 2} \ge \log_5 (x^3 + x^2 + x + 1)$

We can now equate the right-hand side:
$\log_5 x^{3/2\log_5 2} \ge \log_5 (x^3 + x^2 + x + 1)$

This implies:
$x^{3/2\log_5 2} \ge x^3 + x^2 + x + 1$

Testing and analyzing this expression results in no valid $x$ satisfying the inequality within real values since exponential growth and polynomial terms do not align. Thus, the inequality cannot be satisfied, and no solution satisfies the given conditions.

Therefore, the solution to the problem is: No solution.