Logarithm Addition Practice Problems & Rules

To solve this problem, we will use the property of logarithms that allows us to combine the sum of two logarithms:

• Step 1: Identify the formula. We use the property $\log_b(x) + \log_b(y) = \log_b(x \cdot y)$ where both logarithms must have the same base.
• Step 2: Recognize the base. Here, both logarithms are in base 10: $\log_{10}3$ and $\log_{10}4$.
• Step 3: Apply the property. Add the two logarithms using the formula: $\log_{10}3 + \log_{10}4 = \log_{10}(3 \cdot 4)$.
• Step 4: Perform the multiplication. Compute $3 \cdot 4$ to get 12.
• Step 5: Express the result as a single logarithm: $\log_{10}12$.

Therefore, the expression $\log_{10}3 + \log_{10}4$ simplifies to $\log_{10}12$.

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

• Step 1: Identify the given expression as $\log_2 4 + \log_2 5$.
• Step 2: Use the sum of logarithms rule to simplify the expression.
• Step 3: Calculate the product and express the result.

Let's work through each step:

Step 1: We have $\log_2 4 + \log_2 5$ as our expression.

Step 2: Apply the sum of logarithms formula:

$\log_2 4 + \log_2 5 = \log_2 (4 \cdot 5)$

Step 3: Calculate the product:

$4 \times 5 = 20$

Thus, $\log_2 (4 \cdot 5) = \log_2 20$.

Therefore, the solution to the problem is $\log_2 20$.

To solve this problem, we'll apply the following steps:

• Step 1: Identify given logarithms and their base.
• Step 2: Employ the sum of logarithms property to combine the terms.
• Step 3: Calculate the resulting argument of the logarithm.

Now, let's work through each step:

Step 1: We have two logarithms: $\log_9 74$ and $\log_9 \frac{1}{2}$, sharing the base of $9$.

Step 2: Since the bases are the same, we use the sum property of logarithms:

$\log_9 74 + \log_9 \frac{1}{2} = \log_9 (74 \times \frac{1}{2})$.

Step 3: Calculate the product $74 \times \frac{1}{2}$:

$74 \times \frac{1}{2} = 37$.

So, we have:

$\log_9 (74 \times \frac{1}{2}) = \log_9 37$.

Therefore, the solution to the problem is $\log_9 37$.

$2\log_82=\log_82^2=\log_84$

$2\log_82+\log_83=\log_84+\log_83=$

$\log_84\cdot3=\log_812$

Where:

$3\log_49=\log_49^3=\log_4729$

y

$8\log_4\frac{1}{3}=\log_4\left(\frac{1}{3}\right)^8=$

$\log_4\frac{1}{3^8}=\log_4\frac{1}{6561}$

Therefore

$3\log_49+8\log_4\frac{1}{3}=$

$\log_4729+\log_4\frac{1}{6561}$

$\log_ax+\log_ay=\log_axy$

$\left(729\cdot\frac{1}{6561}\right)=\log_4\frac{1}{9}$

$\log_49^{-1}=-\log_49$