How do I solve this exponential equation problem?

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How do I solve this exponential equation problem?

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Answer 1 · 2018-05-17T00:20:19

Please see the explanation.

Explanation:

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

$P (13) = 147 {(1 + 0.0151)}^{13}$ $P(13) = 147(1+0.0151)^13$

$P (13) \approx 178.62 million$ $P(13) ~~ 178.62" million"$

The second question asks for $P (20)$ $P(20)$

$P (20) = 147 {(1 + 0.0151)}^{20}$ $P(20) = 147(1+0.0151)^20$

$P (20) \approx 198.38 million$ $P(20) ~~ 198.38" million"$

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

$2000 = 147 {(1 + 0.0151)}^{t}$ $2000 = 147(1+0.0151)^t$

$\frac{2000}{147} = {(1 + 0.0151)}^{t}$ $2000/147= (1+0.0151)^t$

$ln ({(1 + 0.0151)}^{t}) = \frac{2000}{147}$ $ln((1+0.0151)^t)= 2000/147$

$t = \frac{2000}{147 ln (1 + 0.0151)}$ $t = 2000/(147ln(1+0.0151))$

$t = 907.8$ $t= 907.8$

The year will be $2907$ $2907$ .

Answer 2 · 2018-05-17T00:34:32

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

Explanation:

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

$P (t) = 147 {(1 + .0151)}^{t}$ $P(t) = 147(1+.0151)^t$
$P (t) = 147 {(1.0151)}^{t}$ $P(t) = 147(1.0151)^t$

Plug in values:

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

$P (13) = 147 {(1.0151)}^{13}$ $P(color(blue)(13)) = 147(1.0151)^(color(blue)(13))$

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

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

$P (20) = 147 {(1.0151)}^{20}$ $P(color(blue)(20)) = 147(1.0151)^(color(blue)(20))$

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

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

$2, 000 = 147 {(1.051)}^{t}$ $2,000 = 147(1.051)^t$

${1.051}^{t} = \frac{2, 000}{147}$ $1.051^t = (2,000)/147$

$t = {log}_{1.051} (\frac{2, 000}{147})$ $t = log_(1.051)((2,000)/147)$

$t = 52.48$ $t = 52.48$

$Year = 2000 + 52.48 = 2052$ $"Year " = 2000 + 52.48 = 2052$

The year will be 2052.

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

${log}_{1.051} (\frac{2, 000}{147}) = \frac{{log}_{a} (\frac{2000}{147})}{{log}_{a} (1.051)}$ $log_(1.051)((2,000)/147) = frac{log_a(2000/147)}{log_a(1.051)}$

where $a$ $a$ is whatever base you choose.

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How do I solve this exponential equation problem?

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