Power Property of Logorithms - Examples, Exercises and Solutions

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 inequality $7\log_4 2 < \log_4 x$, we will follow these steps:

• Step 1: Simplify $7\log_4 2$ using the logarithm properties.
• Step 2: Write the inequality $\log_4(2^7) < \log_4 x$.
• Step 3: Solve for $x$ by converting the logarithmic inequality into an exponential form.

Let's now proceed with these steps:

Step 1: Using the power property of logarithms, we have $7\log_4 2 = \log_4 (2^7)$. This step simplifies the multiplication into a single logarithmic term.

Step 2: Using substitution in the inequality, we write it as $\log_4 (2^7) < \log_4 x$.

Step 3: Since logarithms are one-to-one functions, we can conclude that if $\log_4 (2^7) < \log_4 x$, then $2^7 < x$. This results from the property $a^c = a^d \iff c = d$ where the bases are equal.

Therefore, the solution to the inequality is $\mathbf{2^7 < x}$.

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:

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

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 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 $\log_8 10$.

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 $3\log_5 \frac{7}{24}$.