Question

How do I solve this exponential equation problem?

Answer 1

Please see the explanation.

Explanation:

The first question asks for `P(13)`:

`P(13) = 147(1+0.0151)^13`

`P(13) ~~ 178.62" million"`

The second question asks for `P(20)`

`P(20) = 147(1+0.0151)^20`

`P(20) ~~ 198.38" million"`

For the third question, the population equal to 2 billion translates into setting `P(t) = 2000` and then solving for t:

`2000 = 147(1+0.0151)^t`

`2000/147= (1+0.0151)^t`

`ln((1+0.0151)^t)= 2000/147`

`t = 2000/(147ln(1+0.0151))`

`t= 907.8`

The year will be `2907`.

Answer 2

1) `178.621 " million people"`
2) `198.378 " million people"`
3) The year will be 2052.

Explanation:

To emphasize, `P(t)` measures population in millions, and t represents the number of years since 2000.

`P(t) = 147(1+.0151)^t`
`P(t) = 147(1.0151)^t`

Plug in values:

1) The problem asks about the year 2013, which is `color(blue)(13)` years after the year 2000.

`P(color(blue)(13)) = 147(1.0151)^(color(blue)(13))`

`P(13) = 178.620827 " million people"`

2) The year 2020 is `color(blue)(20)` years after the year 2000.

`P(color(blue)(20)) = 147(1.0151)^(color(blue)(20))`

`P(20) = 198.378175 " million people"`

3) Set 2 billion people equal to `P(t)`. Recall that `P(t)` measures in millions. `color(green)(2" billion" = 2,000" million")`

`2,000 = 147(1.051)^t`

`1.051^t = (2,000)/147`

`t = log_(1.051)((2,000)/147)`

`t = 52.48`

`"Year " = 2000 + 52.48 = 2052`

The year will be 2052.

Note: If you have only a scientific calculator that can perform logarithms on only base `e` or base `10`, use the logarithm change of base formula `log_b(x) = log_a(x)/log_a(b)`, so

`log_(1.051)((2,000)/147) = frac{log_a(2000/147)}{log_a(1.051)}`

where `a` is whatever base you choose.