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MATH 180\hfill FALL 1995

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\centerline{ASSIGNMENT \#5}

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1.7, \#2.  $3\ln e+\ln\bigl({1/ e}\bigr)=3\ln e-\ln e=3-1=2$.

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1.7, \#11.  $2e^{3x}=4e^{5x}$ is equivalent to
${1\over2}=e^{(5-3)x}=e^{2x}$.  Thus $x={1\over2}\ln{1\over2}=
-.34657$.

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1.7, \#15.  $P=10e^{0.917t}=10 (2.502)^t$
represents exponential growth since the base $2.502>1$,
or equivalently, since $0.917>0$.

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1.7, \#17.  $P=79e^{-2.5t}=79(.0821)^t$ represents exponential decay
since the base $.0821<1$, or equivalently, since $-2.5<0$.

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1.7, \#27.  This is similar to 1.6, \#9 in Assignment \#4.
Apply the natural logarithm to both sides of $P=P_0e^{kt}$
and solve for $t$:
$$t={1\over k}(\ln  P-\ln P_0).$$

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Comments:  The problems in this assignment are straightforward
exercises in exponential and logarithm functions.
Review the material if your writeup is not perfect.

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