MATH 130 · FRIDAY, OCTOBER 2

From Exponentials to Logarithms

Finish the Section 4.2 models, then learn the logarithm as the exponent

50 minutes · Guided notes handout · Quiz 6 (Section 4.3) on Wednesday, October 7

What the class data says

Section 4.2 · current module

56%

Lowest: continuous interest (31%), graphing $b^x$ with its asymptote (36%), simple vs. compound interest (44%).

Section 4.3 · next module

18%

Lowest: natural-log conversion (8%), solving exponential equations with logs (10–13%), evaluating logs (18%).

Today’s pathway: graph $b^x$ → compare interest models → growth and decay → logs are exponents → evaluate → solve.

Focus · 2 minFriday, Oct 2

Warm-up: exponent fluency

Handout page 1 · 2 minutes
  1. Evaluate $2^{-3}$ and $\left(\tfrac12\right)^{-2}$.
  2. Evaluate $5000(1.04)^3$ to the nearest cent.
  3. Use a calculator: $e^{0.5}\approx$ ?

Retrieve · 3 minFriday, Oct 2

Whiteboard Example 1: graph $f(x)=b^x$

Table, graph, asymptote

$f(x)=2^x$ and $g(x)=\left(\tfrac13\right)^x$

  1. $x=-2,-1,0,1,2$: $f$ gives $\tfrac14,\tfrac12,1,2,4$; $g$ gives $9,3,1,\tfrac13,\tfrac19$.
  2. Both pass through $(0,1)$ because $b^0=1$.
  3. Horizontal asymptote $y=0$. Domain $(-\infty,\infty)$, range $(0,\infty)$.
  4. $b>1$: growth (rises). $0<b<1$: decay (falls).

Model · 5 minFriday, Oct 2

Explore the base $b$

Every curve passes through $(0,1)$ and stays above $y=0$. What happens at $b=1$?

Explore · 1 minFriday, Oct 2

Whiteboard Example 2: three interest models

$\$2000$ at $5\%$ for $6$ years

Find the balance under simple, quarterly, and continuous interest.

  1. Simple: $A=P(1+rt)=2000(1+0.05\cdot6)=\$2600.00$
  2. Quarterly: $A=P\left(1+\tfrac rn\right)^{nt}=2000\left(1+\tfrac{0.05}{4}\right)^{24}\approx\$2694.70$
  3. Continuous: $A=Pe^{rt}=2000e^{0.3}\approx\$2699.72$
  4. Simple interest $<$ compound $<$ continuous, for the same $P$, $r$, $t$.

Convert the rate to a decimal. Compute the exponent $nt$ or $rt$ first, then use parentheses.

Model · 6 minFriday, Oct 2

Whiteboard Example 3: growth, decay, and models

(a) Decay by a percent

A $\$24{,}000$ car loses $15\%$ of its value each year. Find its value after $4$ years.

$V=24000(1-0.15)^4=24000(0.85)^4$

$\approx\$12{,}528.15$

(b) Evaluate a given model

$P(t)=18000(1.023)^t$ gives a town’s population $t$ years after 2020. Estimate the population in 2030.

$t=2030-2020=10$

$P(10)\approx22{,}596$ people

Growth factor $1+r$; decay factor $1-r$. Always find $t$ from the problem’s starting point.

Model · 4 minFriday, Oct 2

SECTION 4.3

A logarithm is an exponent

$\log_b x$ answers the question: “$b$ to what power gives $x$?”

Whiteboard Example 4: convert between forms

$\log_b x = y \iff b^y = x$   ·   $\log x=\log_{10}x$   ·   $\ln x=\log_e x$

$\log_2 32=5$

$\iff 2^5=32$

$10^{-2}=0.01$

$\iff \log 0.01=-2$

$\ln x=3$

$\iff e^3=x$

$e^{0.4}=y$

$\iff \ln y=0.4$

Model · 5 minFriday, Oct 2

Whiteboard Example 5: evaluate without a calculator

$\log_3 81$

$3^{?}=81 \Rightarrow \mathbf{4}$

$\log_5 \tfrac1{25}$

$5^{?}=5^{-2} \Rightarrow \mathbf{-2}$

$\log 1000$

$10^{?}=1000 \Rightarrow \mathbf{3}$

$\ln e^7$

$e^{?}=e^7 \Rightarrow \mathbf{7}$

$\log_4 2$

$4^{1/2}=2 \Rightarrow \mathbf{\tfrac12}$

$\log_9 1$

$9^{0}=1 \Rightarrow \mathbf{0}$

$\log_b x$ is defined only for $x>0$. So $\log_2(-8)$ and $\ln 0$ are undefined.

Model · 4 minFriday, Oct 2

Whiteboard Example 6: solve $\log_b a=c$

$\log_x 49=2$

$x^2=49 \Rightarrow x=7$  (reject $-7$)

$\log_2 x=-3$

$x=2^{-3}=\tfrac18$

$\ln x=2.5$

$x=e^{2.5}\approx12.1825$

$\log_3(2x-1)=2$

$2x-1=3^2=9 \Rightarrow x=5$

Convert to exponential form first. A base must be positive and not equal to $1$.

Model · 5 minFriday, Oct 2

Whiteboard Example 7: calculator and change of base

Round to four decimal places

$\log 250$,   $\ln 0.3$,   $\log_2 20$

  1. $\log 250\approx2.3979$  (use the LOG key)
  2. $\ln 0.3\approx-1.2040$  (negative because $0.3<1$)
  3. $\log_2 20=\dfrac{\ln 20}{\ln 2}\approx4.3219$
  4. Reasonable? $2^4=16$ and $2^5=32$, so the answer is between $4$ and $5$.

Model · 3 minFriday, Oct 2

Whiteboard Example 8: solve exponential equations

(a)

$5^x=42$

$x=\log_5 42=\dfrac{\ln42}{\ln5}$

$x\approx2.3223$

(b)

$3e^{2x}=45$

$e^{2x}=15 \Rightarrow 2x=\ln15$

$x\approx1.3540$

(c) Back to Example 2

When does $\$2000$ at $5\%$ continuous reach $\$4000$?

$2000e^{0.05t}=4000$

$e^{0.05t}=2 \Rightarrow 0.05t=\ln2$

$t\approx13.86$ years

Isolate the exponential expression first. Then take $\ln$ of both sides (or convert to log form).

Model · 6 minFriday, Oct 2

Quiz 6 readiness check

Handout page 4 · no notes · 3 minutes
Show the decisive step
  1. $\$3000$ at $6\%$ compounded continuously for $5$ years. Find $A$.
  2. Write $\ln y=1.8$ in exponential form.
  3. Evaluate $\log_2\tfrac1{16}$.
  4. Solve $2^x=11$. Round to four decimal places.

Assess · 4 minFriday, Oct 2

Choose the next ALEKS target

If item 1 was the miss

Finish Section 4.2: continuous interest, then simple vs. compound interest.

If items 2–4 were the miss

Start Section 4.3 with conversion, then evaluation, then solving.

Quiz 6: Wednesday, October 7 on Section 4.3. Write one specific ALEKS topic on your handout before you leave.

Target · 1 minFriday, Oct 2