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Mathematics 8 Online
smolgaybean06:

okay so i have a word problem that's exponential decay, but I cant figure out what to do. 2. A computer valued at $6500 depreciates at the rate of 14.3% per year. a. Write a function that models the value of the computer. b. Find the value of the computer after three years. c. After how many years will the cost of the computer be $1800? can someone help me?

AZ:

Exponential equations are in the form of y = a(b)^x the initial value is going to be 'a' we're decreasing by 14.3% a year can you write 14.3% as a decimal? the value of 'b' is going to be 1 - r where r is that rate as a decimal so what if 14.3% as a decimal and then what is 1 minus that number

smolgaybean06:

1.43 - 1 then? it'd be - 0.43?

AZ:

no 14.3% % means over 100 so what is 14.3 / 100

smolgaybean06:

0.143?

AZ:

yup and then what is 1 - 0.143 = ? that's the number that 'b' is going to be in y = a(b)^x

AZ:

we subtract 0.143 from 1 because we're decreasing every year if the value of the computer INCREASED each year by 14.3% then we would be ADDING

smolgaybean06:

okay so 0.857 would be 1 - 0.143, the equation would be y = 6500(0.857)^x ?

AZ:

Yup! And then to answer part 2, you plug in x = 3 That will give you the value of the computer in 3 years

smolgaybean06:

so it would be 4091.24? for the part 2 that is

AZ:

Yes

AZ:

Or actually, if you're rounding to the hundredths place then it would be 4091.25

AZ:

For part 3, we want to figure out what 'x' is (that is, how many years it takes) when 'y' (or the price) is at 1800 The original equation is y = 6500(0.857)^x y is 1800 1800 = 6500(0.857)^x To solve for 'x', first divide both sides by 6500

smolgaybean06:

@az wrote:
Or actually, if you're rounding to the hundredths place then it would be 4091.25
yes thank you, I forgot tho round

smolgaybean06:

to*

smolgaybean06:

@az wrote:
For part 3, we want to figure out what 'x' is (that is, how many years it takes) when 'y' (or the price) is at 1800 The original equation is y = 6500(0.857)^x y is 1800 1800 = 6500(0.857)^x To solve for 'x', first divide both sides by 6500
so, 0.28 = (0.857)^x ?

AZ:

so when there's multiple steps, you never ever want to round early you want to keep as many digits until you reach the last step so we have 0.276923 = (0.857)^x do you know about the log rules? \(\log a^b = b ~\log a\) so if we take the log on both sides, we get log 0.276923 = x * log 0.857

AZ:

so now to solve for x, you just have to divide both sides by log 0.857 and you can first use a calculator and find out what log 0.276923 is equal to and what log 0.857 is equal to

smolgaybean06:

okay so log 0.276923 = -0.5576409722 and log 0.857 = -0.06701917808 so now i would just divide both sides by -0.06701917808?

AZ:

yes, and what do you get then?

smolgaybean06:

8.320617891 = x ?

AZ:

you got it

smolgaybean06:

so now it would be 1800 = 6500(0.857)^8.32 ?

AZ:

uhh we just had to solve for x 8.32 is the answer

smolgaybean06:

oh! okay, thank you so much!

AZ:

You're welcome!

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