1. ## Exponential Functions

Investor A invests $1000 at an interest rate of 5% compounded continuously and investor B invests$500 at a rate of 8% compounded continuously. If both of
them invest at the beginning of 2008, when will the value of their investments
be equal.

Kindly show details

2. Originally Posted by Kblack
Investor A invests $1000 at an interest rate of 5% compounded continuously and investor B invests$500 at a rate of 8% compounded continuously. If both of
them invest at the beginning of 2008, when will the value of their investments
be equal.

Kindly show details
$1000(1 + \frac{5}{100})^{n} = 500(1 + \frac{8}{100})^{n}$

$2(1 + \frac{5}{100})^{n} = (1 + \frac{8}{100})^{n}$

$2(\frac{21}{20})^n = (\frac{27}{25})^{n}$

$2^{ \frac{1}{n} } = \frac{27}{25} \div \frac{21}{20}$

$2^{ \frac{1}{n} } = \frac{36}{35}$

$\frac{1}{n} = \log_{2} \frac{36}{35}$

I get $n = 24,605$

So I'd say between 24 and 25 years.

3. Originally Posted by Kblack
Investor A invests $1000 at an interest rate of 5% compounded continuously and investor B invests$500 at a rate of 8% compounded continuously. If both of
them invest at the beginning of 2008, when will the value of their investments
be equal.

Kindly show details
$
1000(1.05)^n = 500(1.08)^n \implies 2 = \left( \frac{1.08}{1.05} \right)^n
$

$\implies n = \dfrac{1}{\log_2 \left( \frac{1.08}{1.05} \right)} = \dfrac{\ln 2}{\ln \left( \frac{1.08}{1.05} \right)} \approx 24.6050977$