Answer:
42330
Step-by-step explanation:
1/5 of x is 8466
multiply is by 5 to cancel out the 1/5
5 times 6466 is 42330
Answer:
Choice A
Step-by-step explanation:
The scenario presented relates to exponential growth models; the population of bacteria is growing at an exponential rate given by the equation;
In this case B represents the population of the bacteria, t the time in minutes, k the growth constant and 1000 represents the initial population at time 0.
After 15 minutes, the population of bacteria grows to 8520. This implies that B is 8520 while t is 15. We substitute this values into the given equation and solve for k, the growth constant;
Divide both sides by 1000;
The next step is to introduce natural logs on both sides of the equation;
Answer:
Step-by-step explanation:
Given:
To find :
Solution:
In order to find the logarithm of 162 in terms of p and q, we can use the properties of logarithms.
We can start by expressing 162 as a product of prime factors:
Now, we can use the properties of logarithms to simplify this expression:
Since log(ab) = log(a) + log(b), we can split this into separate logarithms:
Now, we can use the fact that q = log 3:
Using the property, we get:
Now, substitute the values of p and q:
So, the logarithm of 162 in termsof p and q is:
Answer:
log 162 = 6p + 2q
Step-by-step explanation:
To write log 162 in terms of p and q, we can use the following steps:
- First, we can write 162 as a product of powers of 2 and 3, such as 162 = 2 x 3^4.
- Next, we can use the property of logarithms that log ab = log a + log b to write log 162 = log 2 + log 3^4.
- Then, we can use another property of logarithms that log a^n = n log a to write log 3^4 = 4 log 3.
- Finally, we can substitute p = log 2 and q = log 3 to get log 162 = p + 4q.
We can write 162 as follows:
```
162 = 2^6 * 3^2
```
Therefore,
```
log 162 = log (2^6 * 3^2)
```
Using the logarithmic properties of addition and multiplication, we can simplify this to:
```
log 162 = 6 * log 2 + 2 * log 3
```
Finally, substituting p = log 2 and q = log 3, we get the following expression:
```
log 162 = 6p + 2q
```
Therefore, log 162 can be written as **6p + 2q** in terms of p and q.
Okay, let's break this down step-by-step:
* log 162 = log (2^4 * 3^2) (by prime factorization)
* log (2^4 * 3^2) = 4log2 + 2log3 (by properties of logarithms)
* Let p = log 2 and q = log 3
* Substituting:
* log 162 = 4p + 2q
Therefore, log 162 can be written as 4p + 2q, where p = log 2 and q = log 3.
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To express log 162 in terms of p (log 2) and q (log 3), you can use logarithm properties, particularly the change of base formula. The change of base formula states that:
log_b(a) = log_c(a) / log_c(b)
In your case, you want to find log 162:
log 162 = log 2^1 * 3^4
Now, we can use the change of base formula with base 10 (or any other base):
log 162 = (log 2^1 * 3^4) / (log 10)
Since log 10 is simply 1 (logarithm of 10 to any base is 1), we can simplify further:
log 162 = (log 2^1 * 3^4) / 1
Now, apply the properties of logarithms to split the logarithm of a product into a sum of logarithms:
log 162 = (log 2^1) + (log 3^4)
Now, we can replace log 2 with p and log 3 with q:
log 162 = p + (4q)
So, log 162 in terms of p and q is:
log 162 = p + 4q
To write log 162 in terms of p and q, we can use the following steps:
- First, we can write 162 as a product of powers of 2 and 3, such as 162 = 2 x 3^4.
- Next, we can use the property of logarithms that log ab = log a + log b to write log 162 = log 2 + log 3^4.
- Then, we can use another property of logarithms that log a^n = n log a to write log 3^4 = 4 log 3.
- Finally, we can substitute p = log 2 and q = log 3 to get log 162 = p + 4q.