exponential growth. how would I find x?

Question

Given y=(0.8)(0.84)^x

x= time period in years

when will y= 1,000,000 in terms of x?


*can you show it step by step please =]?

Answer 1

use logarithms

y=(0.8)* (0.84)^x
log y = log (0.8) + x log(0.84)

log (1,000,000) = log (0.8) + x log(0.84)

6 = -0.097 + x * (-0.076)

x= -80.2 years

So, check the original problem! Since the (0.84)<1, any positive exponent x greater than 1 will always give a number less than 1. You will NEVER get to 1,000,000 in positive years with that expression!

Answer 2

We want to know what x is when y = 1,000,000...so let's set y = 1,000,000 in the equation and solve for x.

1000000 = (0.8)(0.84)^x

Now divide both sides by 0.8.

1250000 = (0.84)^x

Now let's take the natural logarithm of both sides:

ln (1250000) = ln ((0.84)^x)

Remember that ln (a^b) = b ln a, so we can bring the x out front:

ln (1250000) = x ln (0.84)

Now divide by ln(0.84):

ln (1250000) / ln (0.84) = x

Using a calculator, x is about -80.5184.

Answer 3

I will tell you the steps, but you will have to "show" them to yourself on paper =)

First, divide both sides by 0.8. Then take the logarithm (either base-10 or natural, doesn't matter) of both sides. Finally, divide both sides by log(0.84) to isolate x on one side of the equation.

Hope that helps!

Answer 4

let us rearrange the equation

y/0.8=0.84^x


apply logon both sides log(base0.84)(y/0.8)=log(base0.84)(0.84)^x
==>log(base0.84)(y/0.8)=x
put y=1000000 we get log(base0.84)(125000)=x

now since it is not easy to find log of a number with base like 0.84

we will use following properties of a log
log(basea)b
=log(basec)b/log(basec)a
where c is any other number so our equation can be written as
log(base 10)125000/log(base10)0.84=x

use a calculator/log table now to get the answer

Answer 5

This isn't exponential growth, it is exponential decay.
1,000,000=088*084^x
.84^x=1250000
x ln .84=ln 1250000
x=ln 1250000/ ln .84=-80.5