What value correctly evaluates log7492+log77

Question

using the properties of logarithms?(1 point)
Responses

3
3

6
6

5
5

4

Answer 1

To evaluate ${\log }_7(492)+{\log }_7(7)$, we can use the property of logarithms which states that ${\log }_b(m)+{\log }_b(n)={\log }_b(m\cdot n)$.

Applying this property here:

$${\log }_7(492)+{\log }_7(7)={\log }_7(492\cdot 7)$$

Calculating $492\cdot 7$:

$$492\cdot 7=3444$$

Now we have:

$${\log }_7(492)+{\log }_7(7)={\log }_7(3444)$$

Next, we need to determine what power of $7$ gives $3444$. However, we can simplify this process. The value of ${\log }_7(7)$ is $1$ since $7^1=7$.

Thus:

$${\log }_7(492)+1$$

To compare with the given answers, we can estimate ${\log }_7(492)$. Since $7^2=49$ and $7^3=343$, we know that $7^4=2401$ and $7^5=16807$, so ${\log }_7(492)$ will be slightly above $3$ because $343<492<2401$.

Since ${\log }_7(492)+1$ is also slightly above $4$:

• ${\log }_7(492)\approx 3$
• ${\log }_7(492)+1\approx 4$

Therefore, the value that correctly evaluates ${\log }_7(492)+{\log }_7(7)$ is:

4.