How do I solve this exponential equation problem?

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2 Answers
May 17, 2018

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.

May 17, 2018

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.