MATH 130 · Guided notes · Sections 4.2 and 4.3
Name ________________________________ Fill in each example as it is worked on the whiteboard.
Class ALEKS progress: Section 4.2 is at 56% and Section 4.3 is at 18%. The lowest skills are continuous interest, graphing $b^x$ with its asymptote, converting natural logarithms, and solving exponential equations with logarithms. Today targets exactly those skills.
1. Evaluate $2^{-3}$ and $\left(\tfrac12\right)^{-2}$.
$\tfrac18$ and $4$
2. Evaluate $5000(1.04)^3$ to the nearest cent.
$\$5624.32$
3. Use a calculator: $e^{0.5}\approx$ ______
$1.6487$
An exponential function has the form $f(x)=b^x$ where $b>0$ and $b\ne$ $1$.
y-intercept: $(0,1)$
Horizontal asymptote: $y=0$
Domain: $(-\infty,\infty)$
Range: $(0,\infty)$
If $b>1$ the graph shows growth; if $0<b<1$ the graph shows decay.
Complete the table, plot both functions on the grid, and draw the asymptote as a dashed line: $f(x)=2^x$ and $g(x)=\left(\tfrac13\right)^x$.
| $x$ | $-2$ | $-1$ | $0$ | $1$ | $2$ |
|---|---|---|---|---|---|
| $f(x)=2^x$ | $\tfrac14$ | $\tfrac12$ | $1$ | $2$ | $4$ |
| $g(x)=\left(\tfrac13\right)^x$ | $9$ | $3$ | $1$ | $\tfrac13$ | $\tfrac19$ |
Which function is growth? $f(x)=2^x$ Which is decay? $g(x)=\left(\tfrac13\right)^x$
For $h(x)=3^x$, find $h(-2)$, $h(0)$, and $h(2)$. State the asymptote and whether $h$ shows growth or decay.
$h(-2)=\tfrac19$, $h(0)=1$, $h(2)=9$. Asymptote $y=0$. Growth, since $3>1$.
$P$ = principal, $r$ = annual rate as a decimal, $t$ = years, $n$ = compounding periods per year.
| Model | Formula | On Test 2? |
|---|---|---|
| Simple interest | $A=P(1+rt)$ | Not printed. Memorize. |
| Compound interest | $A=P\left(1+\tfrac rn\right)^{nt}$ | Printed on the test. |
| Continuous compounding | $A=Pe^{rt}$ | Printed on the test. |
Common values of $n$: annually $1$, quarterly $4$, monthly $12$, daily $365$.
$\$2000$ is invested at $5\%$ for $6$ years. Find the balance with (a) simple interest, (b) interest compounded quarterly, and (c) interest compounded continuously.
(a) $A=2000(1+0.05\cdot6)=2000(1.3)=\$2600.00$
(b) $A=2000\left(1+\tfrac{0.05}{4}\right)^{4\cdot6}=2000(1.0125)^{24}\approx\$2694.70$
(c) $A=2000e^{0.05\cdot6}=2000e^{0.3}\approx\$2699.72$
Order: simple $<$ quarterly $<$ continuous. Round only at the final step.
$\$5000$ is invested at $4\%$ for $10$ years. Find the balance if interest is compounded monthly, and if it is compounded continuously.
Monthly: $5000\left(1+\tfrac{0.04}{12}\right)^{120}\approx\$7454.16$ Continuous: $5000e^{0.4}\approx\$7459.12$
Growth by $r$ per period: multiply by $1+r$ Decay by $r$ per period: multiply by $1-r$
(a) A $\$24{,}000$ car loses $15\%$ of its value each year. Find its value after $4$ years.
$V=24000(1-0.15)^4=24000(0.85)^4$
$V\approx\$12{,}528.15$
(b) $P(t)=18000(1.023)^t$ gives a town’s population $t$ years after 2020. Estimate the population in 2030.
$t=2030-2020=10$
$P(10)=18000(1.023)^{10}\approx22{,}596$ people
A $60$ mg dose of medicine decreases by $12\%$ each hour. How much remains after $5$ hours?
$60(1-0.12)^5=60(0.88)^5\approx31.66$ mg
Key idea: a logarithm is an exponent. $\log_b x$ answers the question “$b$ to what power gives $x$?”
$\log_b x = y \quad\iff\quad$ $b^y = x$
$\log x$ means base $10$ (common log)
$\ln x$ means base $e$ (natural log)
Restrictions: the base satisfies $b>0$, $b\ne1$, and the input satisfies $x$ $>0$.
Rewrite each equation in the other form.
| Given | Other form |
|---|---|
| $\log_2 32=5$ | $2^5=32$ |
| $10^{-2}=0.01$ | $\log 0.01=-2$ |
| $\ln x=3$ | $e^3=x$ |
| $e^{0.4}=y$ | $\ln y=0.4$ |
(a) $\log_6 36=2$ $6^2=36$
(b) $10^{1.5}=x$ $\log x=1.5$
(c) $\ln 8=x$ $e^x=8$
(d) $e^{-1}=y$ $\ln y=-1$
Evaluate without a calculator. Ask: “the base to what power gives the input?”
$\log_3 81=$ $4$
$\log_5 \tfrac1{25}=$ $-2$
$\log 1000=$ $3$
$\ln e^7=$ $7$
$\log_4 2=$ $\tfrac12$
$\log_9 1=$ $0$
Why are $\log_2(-8)$ and $\ln 0$ undefined? $b^y$ is always positive, so the input must be positive.
$\log_2 64=$ $6$
$\log 0.001=$ $-3$
$\log_7 \tfrac17=$ $-1$
$\ln\sqrt e=$ $\tfrac12$
$\log_{2} 1=$ $0$
$\log_{8} 2=$ $\tfrac13$
Change of base: $\log_b a=\dfrac{\ln a}{\ln b}$ (or $\dfrac{\log a}{\log b}$). Round to four decimal places.
$\log 250\approx$ $2.3979$
$\ln 0.3\approx$ $-1.2040$
$\log_2 20\approx$ $4.3219$
Reasonableness check for $\log_2 20$: $2^4=$ $16$ and $2^5=$ $32$, so the answer lies between $4$ and $5$. Your turn: $\log_3 50\approx$ $3.5609$
Solve. First convert to exponential form.
(a) $\log_x 49=2$
$x^2=49$, so $x=7$ (reject $-7$: a base must be positive)
(b) $\log_2 x=-3$
$x=2^{-3}=\tfrac18$
(c) $\ln x=2.5$
$x=e^{2.5}\approx12.1825$
(d) $\log_3(2x-1)=2$
$2x-1=3^2=9$, so $x=5$
(a) $\log_x 81=4$ $x=3$
(b) $\log_5 x=3$ $x=125$
(c) $\ln x=-1.2$ $x=e^{-1.2}\approx0.3012$
(d) $\log(x+4)=2$ $x+4=100$, so $x=96$
Solve. Isolate the exponential expression first, then use logarithms. Round to four decimal places.
(a) $5^x=42$
$x=\dfrac{\ln42}{\ln5}\approx2.3223$
(b) $3e^{2x}=45$
$e^{2x}=15$
$2x=\ln15$
$x\approx1.3540$
(c) When does $\$2000$ at $5\%$ compounded continuously reach $\$4000$?
$e^{0.05t}=2$, $t=\tfrac{\ln2}{0.05}\approx13.86$ yr
(a) $4^x=30$ $x=\tfrac{\ln30}{\ln4}\approx2.4534$
(b) $6e^{0.3x}=90$ $e^{0.3x}=15$, $x=\tfrac{\ln15}{0.3}\approx9.0268$
1. $\$3000$ at $6\%$ compounded continuously for $5$ years. Find $A$. ✓ ~ ✗
$3000e^{0.3}\approx\$4049.58$
2. Write $\ln y=1.8$ in exponential form. ✓ ~ ✗
$y=e^{1.8}$
3. Evaluate $\log_2\tfrac1{16}$. ✓ ~ ✗
$-4$
4. Solve $2^x=11$ (four decimal places). ✓ ~ ✗
$x=\tfrac{\ln11}{\ln2}\approx3.4594$
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