Inverting Log Functions Practice - Base Change Problems

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

• Step 1: Identify the property of logarithms that relates inverses.
• Step 2: Apply this property to the given expression.
• Step 3: Compare with provided choices to identify the correct option.

Now, let's work through each step:

Step 1: The problem asks us to find the expression equal to $\frac{1}{\log_4 9}$.

Step 2: We use the logarithmic property $\log_b a = \frac{1}{\log_a b}$. Thus, replacing $b$ with 9 and $a$ with 4, we have:

$\frac{1}{\log_4 9} = \log_9 4$.

Step 3: Comparing this result to the provided choices, we find that the correct answer is $\log_9 4$, corresponding to Choice 1.

Therefore, the solution to the problem is $\log_9 4$.

To solve the problem, follow these steps:

• Step 1: Identify the expression to simplify, $\frac{1}{\ln 8}$.
• Step 2: Apply the change of base formula:

Using the change of base formula, we have:
$\log_8 e = \frac{\ln e}{\ln 8}$ Since $\ln e = 1$, substituting gives:
$\log_8 e = \frac{1}{\ln 8}$

Therefore, the expression $\frac{1}{\ln 8}$ can be rewritten as $\log_8 e$.

Thus, the correct choice is $\log_8 e$.

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 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 $\pm 2$.