Question

A population of 1500 is to decrease 12% in a year. After how many years will the population be 1200?
The answer is 2 years but I NEED help on how to get the answer.
Using the formula y=c*a^x, where y is the new population, c is the old pop, a is the percet, and x is the time.

Answer 1

So, you know the answer, but you don't know how to get to it?

Answer 2

Yes

Answer 3

Looking at the formula you gave, can you tell me what the new population is?

What is the old population?

Can you express the percentage they gave us as a decimal? (This would be the value of a)

And we'll need to solve for time.

Answer 4

1200=1500*.12^x

Answer 5

Great!

Now, we have to divide both sides by 1500!

Can you do that and tell me what you get?

Answer 6

.8 = .12^x

Answer 7

Now, have you learned about logarithms?

Answer 8

Yes

Answer 9

Great!

We should know that:

\(\Large\bf y = b^x\)

\(\Large\bf log_b y = x\)

are both the same things.

Now, for your question, we want to take the log of both sides.

Now, we're solving for x, so we're going to take the log with base .12 on both sides because:

\(\Large\bf log_a b^c = c ~log_a b\)

Answer 10

So we have:

.8 = .12^x

Let's take \( log_{1.2}\) on both sides and we get:

\(\Large\bf log_{1.2} 0.8 = log_{1.2}1.2^x\)

And you should know that:

\(\Large\bf log_b b^a = a\)

So, simplifying what we had earlier we get:

\(\Large\bf log_{1.2} 0.8 = x\)


Now do you know the change of base formula?

And are you able to follow along so far?

Answer 11

Where do you get log1.2?

Answer 12

we did log1.2 on both sides so that way we could isolate x and so that way we can solve for it

Answer 13

Where did the 1.2 part come from though?

Answer 14

Because we had 1.2^x

and we want to get x out of the exponents, so we did:

\(\Large\bf log_{1.2} 1.2^x = x\)

Answer 15

since its decreasing by 12% isnt it supposed to be .88?

Answer 16

Oh, yes! I totally missed that. Good catch!

Answer 17

So, at the end, we should get:

\(\Large\bf log_{0.88} 0.8 = x\)

Also, somewhere in the middle, I accidentally had switched from 0.12 to 1.2, my bad.

Answer 18

Here's the last formula you need to use to solve for x

\(\LARGE\bf log_a x = \frac{log_b x}{log_b a}\)

and so to answer your question, we have:

\(\LARGE\bf log_{0.88} 0.8 = \frac{log_{10} 0.8}{log_{10} 0.88} =\frac{log0.8}{log0.88}\approx 1.75\)

Which is around 2 years.

Answer 19

Oh okay. Thank you so much!

Answer 20

It was my pleasure!