Quiz 11 (Solution)

Date: MATH 1090-4 - Fall 2003

1. If $ 1000 is invested for $x$ years at 8%, compounded quarterly, the future value is given by

   $$S(x) = 1000\cdot (1.02)^{4x}.$$

   
   
   What amount will result in 10 years?

   We use the formula with $x=10$:

   $$S(10) = 1000\cdot (1.02)^{4\cdot 10} = 1000\cdot (1.02)^{40} = 1000 \cdot 2.2080397 = 2208.0397.$$

   
   
   Thus, the future value is (rounded to cents) $2208.04.

2. Find $x$ such that $\log_8(x) = -\frac{1}{3}$.

   For logarithms we have $\log_b(x) = y$ means that $b^y =x$. In this case, then

   $$x = 8^{-\frac{1}{3}} = \frac{1}{8^{\frac{1}{3}}} = \frac{1}{\sqrt[3]{8}} = \frac{1}{2}.$$

   
   
   That is, $x=\frac{1}{2}$.

3. Use your calculator and the change of base formula to find $\log_3(100)$.

   The change of base formula says that, for instance,

   $$\log_b(x) =\frac{\ln(x)}{\ln(b)}.$$

   
   
   So, for $b=3$ we have

   $$\log_3(100) = \frac{\ln(100)}{\ln(3)} = \frac{4.605170186}{1.098612289} = 4.191806548.$$

   
   

4. Find $x$ so that $30=e^{2x}$.

   We take $\ln$ on both sides:

   $$\begin{split}30&=e^{2x}\\ \ln(30) &= \ln(e^{2x}) = 2x\\ \frac{\ln(30)}{2} &=x\\ 1.700598691 &=x. \end{split}$$