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MATH 180\hfill FALL 1995

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%10:00 SECTION\hfill FONG

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\centerline{ASSIGNMENT \#4}

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1.4, \#18.
We proceed as in 1.4, \#17 in the last assignment.
The functions are an exponential, a quadratic, and a cubic.
For $g(x)$, the ratios of consecutive values from the table are
$${3.74\over3.12}={4.49\over3.74}=
  {5.39\over4.49}={6.47\over5.39}=
  {7.76\over6.47}=1.20$$
up to 2 decimal places.  So $g(x)$ is exponential with base $(1.2)^2
=1.44$.  The base is 1.44 and not 1.2 since the inputs
$5.0$, $5.5$, $6.0$, $6.5,\ldots$ change by $.5$.
Indeed, if $g(x)=ab^x$ and
$x$ and $x+\Delta x$ are two consecutive inputs, then
$${g(x+\Delta x)\over g(x)}={ab^{x+\Delta x}\over ab^x}=
  b^{\Delta x}.$$
In the problem $\Delta x=.5$, $b^{(.5)}=1.2$, and $b=1.44$.
The remaining two functions are a
quadratic and a cubic power function.  Since
$${f(9.0)\over f(8.4)}={7.29\over5.93}=1.23,\quad
    \left({9.0\over8.4}\right)^3=1.23,$$
$f(x)$ is cubic and $k(x)$ is necessarily quadratic.

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1.6, \#2, 3.
The calculator or computer should show
the graph of $y=\log 10^x$ as the line $y=x$, and
the graph of $y=10^{\log x}$ as the half-line $y=x$ with $x>0$.
These are different graphs
even though $\log x$ and $10^x$ are inverses functions of each other.
The reason is that
$x$ can be any real number in $\log 10^x$,
whereas $x$ is necessarily positive in $10^{\log x}$
since the logarithm function is defined only for positive arguments.

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1.6, \#5.  $2=(1.02)^t$ is equivalent to $t(\log 1.02)=\log 2$
or $t=35$.

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1.6, \#6.  $7\cdot 3^t=5\cdot 2^t$ is equivalent to
$\log 7+t(\log 3)=\log 5+t(\log2)$.  Thus
$$t={\log 5-\log 7\over \log3-\log2}=-.83.$$

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1.6, \#9.  $P=P_0a^t$ is equivalent to $\log P=\log P_0+t(\log a)$
or
$$t={\log P-\log P_0\over \log a}.$$


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Comments:  The table of values of the exponential function $ab^x$ in
1.4, \#18 differs from the tables in 1.3, \#9 and 1.4, \#17.
Namely, consecutive inputs differ by $\Delta x$, a number unequal
to 1.  The ratio of consecutive values is
$b^{\Delta x}$ rather than $b$.


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