Algebra II: Exponentials & Logarithms

How to do it:

1. Rewrite $y=b^x$ as $x=\log_b y\text{.}$

2. Cancel: $\log_b(b^x)=x$ and $b^{\log_b x}=x$ (for $x>0\text{).}$

3. Graph $y=b^x$ and $y=\log_b x$ as reflections across $y=x\text{.}$

Worked example: $\log_2 8=3$ because $2^3=8\text{.}$ See logarithm definition.

Check: $\log 100$ on a calculator is base 10, so 2. $\ln$ is base $e\text{.}$

How to do it:

1. Products become sums: $\log(xy)=\log x+\log y\text{.}$

2. Quotients become differences: $\log(x/y)=\log x-\log y\text{.}$

3. Exponents come out front: $\log(x^k)=k\log x\text{.}$

4. Change of base: $\log_b a=\dfrac{\log_c a}{\log_c b}$ so a calculator can evaluate any base.

Worked example: $\log_2 24 = \log_2(8\cdot 3)=3+\log_2 3\text{.}$ Replay product rule.

Check: $\log(x+y)$ is not $\log x + \log y\text{.}$ The argument of a real log must stay positive.

How to do it:

1. Isolate the exponential.

2. Take log of both sides (any matching base, or $\ln\text{).}$

3. Use $\log(b^{ct})=ct\log b\text{,}$ then divide to get $t\text{.}$

Worked example: $5\cdot 2^{t}=80\text{.}$ Divide: $2^{t}=16\text{,}$ so $t=4\text{.}$ If the number is not a nice power, $t=\dfrac{\ln 16}{\ln 2}\text{.}$

Check: plug $t$ back into the original. Exponent properties for rationals ($8^{2/3}=4\text{)}$ live in rational exponent as a radical.

How to do it:

1. In $y=ab^t\text{,}$ $a$ is the starting amount and $b$ is the growth factor per step.

2. Growth by $r$ percent means $b=1+r\text{;}$ decay by $r$ percent means $b=1-r\text{.}$ Half-life means $b=\tfrac{1}{2}$ per half-life.

3. Count the steps carefully: $t$ is measured in the same unit the factor uses.

Worked example: $\$500$ at $6\%$ per year for $3$ years: $500(1.06)^3\approx 595.51$ dollars. An $80$-gram sample with a half-life of $5$ days, after $15$ days: that is $3$ half-lives, so $80\left(\tfrac{1}{2}\right)^3=10$ grams.

Check: $b>1$ grows and $0<b<1$ decays. A common slip is writing $b=0.06$ for $6\%$ growth; the factor is $1.06\text{,}$ because you keep the whole and add the interest.

How to do it:

1. Read $b^{m/n}$ as $\left(\sqrt[n]{b}\right)^m\text{:}$ the denominator is the root, the numerator is the power.

2. Take the root first when you can; it keeps the numbers small.

3. A negative exponent means reciprocal, not a negative answer: $b^{-m/n}=\dfrac{1}{b^{m/n}}\text{.}$

Worked example: $8^{2/3}=\left(\sqrt[3]{8}\right)^2=2^2=4\text{.}$ $16^{3/4}=\left(\sqrt[4]{16}\right)^3=2^3=8\text{.}$ $27^{-1/3}=\dfrac{1}{\sqrt[3]{27}}=\dfrac{1}{3}\text{.}$

Check: the exponent rules still hold: $8^{2/3}\cdot 8^{1/3}=8^{1}=8\text{,}$ and $4\cdot 2=8$ agrees. Replay rational exponent as a radical.

How to do it:

1. Isolate the exponential: divide by $a\text{.}$

2. Take $\ln$ (or $\log\text{)}$ of both sides and bring the exponent down: $t\ln b=\ln(d/a)\text{.}$

3. Divide, evaluate, and say what the number means in the story.

Worked example: when does $\$200$ growing $8\%$ per year reach $\$1{,}000\text{?}$ $200(1.08)^t=1000\text{,}$ so $1.08^t=5$ and $t=\dfrac{\ln 5}{\ln 1.08}\approx 20.9\text{.}$ About $21$ years.

Check: $200(1.08)^{20.9}\approx 1000\text{.}$ If the money had only doubled ($t\approx 9$ years) it would be $\$400\text{,}$ so about $21$ years for five times is reasonable.

How to do it:

1. When growth compounds continuously (every instant), the model is $A=Pe^{rt}\text{:}$ $P$ is the start, $r$ the rate as a decimal, $t$ the time.

2. $e\approx 2.718$ is just a number, so $e^{rt}$ is evaluated on a calculator like any other power.

3. To solve for $t\text{,}$ take $\ln$ of both sides; $\ln$ undoes $e\text{,}$ so $\ln e^{rt}=rt\text{.}$

Worked example: $\$1{,}000$ at $5\%$ for $10$ years. Continuous: $1000e^{0.05\cdot 10}=1000e^{0.5}\approx 1648.72\text{.}$ Compounded once a year instead: $1000(1.05)^{10}\approx 1628.89\text{,}$ a little less, because it compounds less often. Doubling time: $e^{0.05t}=2\text{,}$ so $0.05t=\ln 2$ and $t=\dfrac{\ln 2}{0.05}\approx 13.9$ years.

Check: $\ln 2\approx 0.693\text{,}$ and the rule of $70$ says doubling takes about $70/5=14$ years, so $13.9$ is right. More frequent compounding always gives a slightly larger amount, never a smaller one.

How to do it:

1. $y=b^x+k$ shifts up $k\text{;}$ the horizontal asymptote moves to $y=k\text{.}$

2. $y=b^{x-h}$ shifts right $h$ (the sign inside works backwards). $y=-b^x$ reflects across the $x$-axis; $y=c\cdot b^x$ stretches by $c\text{.}$

3. The same rules move $y=\log_b x\text{;}$ there the vertical asymptote $x=0$ shifts with $h\text{.}$

Worked example: $y=3\cdot 2^{x}-4\text{:}$ stretch by $3\text{,}$ shift down $4\text{.}$ Asymptote $y=-4\text{,}$ $y$-intercept $3\cdot 1-4=-1\text{,}$ so $(0,-1)\text{.}$ $y=\log_2(x-1)$ shifts the log right $1\text{,}$ with asymptote $x=1$ and domain $x>1\text{.}$

Check: the horizontal shift is the opposite of the sign you see: $x-1$ moves right. Test the intercept by substituting $x=0\text{.}$

How to do it:

1. Swap $x$ and $y\text{.}$

2. Isolate the exponential, then rewrite it as a log to free $y\text{.}$

3. Check a point: if $(0,3)$ is on the original, $(3,0)$ must be on the inverse.

Worked example: $y=3\cdot 2^{x}\text{.}$ Swap: $x=3\cdot 2^{y}\text{.}$ Divide: $2^{y}=\dfrac{x}{3}\text{.}$ Rewrite: $y=\log_2\dfrac{x}{3}\text{.}$ The point $(0,3)$ on the original becomes $(3,0)\text{:}$ $\log_2\dfrac{3}{3}=0\text{.}$

Check: the inverse of a growth function takes an amount and returns the time, which is exactly what solving for $t$ did above. Domains swap too: the inverse only accepts $x>0\text{.}$