Why does the base of an exponent have to be greater then 1

1 answer:

7 0

It's not 1 that is exactly the problem. It is also 0. And perhaps - 1 although that is a lot trickier than the other two.

Let's start with one. It give 1^1 = 1 for a power of 1. It gives 1^2 = 1 * 1 for a power of 2. It will give 1^100000 = 1 for a power of 100000.

So the base can be 1 but it is known that any power won't change it's value.

The same can be said of 0. 0^1000 = 0^1 = 0. 0^0 is really a bad dude. You won't learn about that for awhile.

(-1)^11 is different from (-1)^10. The first gives -1 and the second gives 1. So you have to be careful with -1.

There is nothing wrong with (1/4)^3. The power of three and the base of 1/4 are fine. The answer is (1/4)*(1/4) * (1/4) = 1/64 which is perfectly good answer.

It works for negatives too. (-1/2)^3 = - 1/8

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The formula Upper A equals 23.1 e Superscript 0.0152 tA=23.1e0.0152t models the population of a US state, A, in millions, t ye

Wewaii [24]

Answer:

a) $A(t=0)= 23.1 e^{0.0152(0)}=23.1e^0 =23.1$

b) $t = \frac{ln(\frac{28.3}{23.1})}{0.0152}=13.357 years$

So for this case the answer would be 13.347 years, the population will be 28.3 million and ye year would be 2000+13.347 and that would be approximately in 2014

Step-by-step explanation:

For this case we assume the following model:

$A(t)= 23.1 e^{0.0152 t}$

Where t is the number of years after 2000/

Part a

For this case we want the population for 2000 and on this case the value of t=0 since we have 0 years after 2000. If we rpelace into the model we got:

$A(t=0)= 23.1 e^{0.0152(0)}=23.1e^0 =23.1$

So then the initial population at year 2000 is 23.1 million of people.

Part b

For this case we want to find the time t whn the population is 28.3 million.

So we need to solve this equation:

$28.3= 23.1 e^{0.0152(t)}$

We can divide both sides by 23.1 and we got:

$\frac{28.3}{23.1}= e^{0.0152t}$

Now we can apply natural log on both sides and we got:

$ln(\frac{28.3}{23.1})= 0.0152 t$

And then for t we got:

$t = \frac{ln(\frac{28.3}{23.1})}{0.0152}=13.357 years$

So for this case the answer would be 13.347 years, the population will be 28.3 million and ye year would be 2000+13.347 and that would be approximately in 2014

3 0

3 years ago

Please answer this question in the picture!

Brums [2.3K]

Answer:

c $113\:m$