Exponential and Logarithmic Functions Study Guide

Fundamentals of Writing Exponential Equations from Data Tables

Exponential functions are mathematically represented by the general equation $f(x) = a \times (b)^{x}$. In this structure, the variable $a$ represents the initial amount (the value of $y$ when $x = 0$), and the variable $b$ represents the growth or decay factor (the base). To determine if a table represents growth or decay, one must observe the change in the output value ($y$) as the input value ($x$) increases. If the $y$-values increase, the table represents growth ($b > 1$); if they decrease, it represents decay ($0 < b < 1$).

In the first example table, the coordinates provided are $(0, 2)$, $(1, 8)$, $(2, 32)$, and $(3, 128)$. By identifying the value at $x = 0$, we find the initial amount $a = 2$. To find the growth factor $b$, we calculate the ratio between successive $y$-values: $\frac{8}{2} = 4$. Since the values are increasing, this is a growth function. The final exponential equation is $f(x) = 2 \times 4^{x}$.

In the second example table, the coordinates are $(0, 5)$, $(1, 2.5)$, $(2, 1.25)$, and $(3, 0.625)$. Here, the initial amount $a = 5$. The factor $b$ is determined by the ratio $\frac{2.5}{5} = 0.5$. Because the $y$-values are decreasing, this represents decay. The resulting equation is $f(x) = 5 \times 0.5^{x}$.

The third example provides data starting at $x = 2$: $(2, 80)$, $(3, 160)$, $(4, 320)$, and $(5, 640)$. First, the growth factor is identified by $\frac{160}{80} = 2$. To find the initial amount $a$, we must move backward to $x = 0$. If at $x = 2$, $y = 80$, then at $x = 1$, $y = 40$, and at $x = 0$, $y = 20$. Thus, $a = 20$. The equation is $f(x) = 20 \times 2^{x}$.

The fourth example contains coordinates $(1, 2)$, $(2, 1)$, $(3, 0.5)$, $(4, 0.25)$, and $(5, 0.25)$. Note that the last value appears to be a typo in the transcript and should be $0.125$ to maintain the ratio. The decay factor is $\frac{1}{2} = 0.5$. Moving back from $x = 1$ where $y = 2$, we find that at $x = 0$, $y = 4$. Therefore, $a = 4$ and the equation is $f(x) = 4 \times 0.5^{x}$.

Constructing Exponential Functions from Real-World Scenarios

When writing exponential functions from word problems, the formula can be expressed either as $f(x) = a \times b^{x}$ or $f(x) = a(1 \pm r)^{x}$, where $r$ is the rate of increase or decrease expressed as a decimal.

In the scenario regarding the Amazon CEO, he makes approximately $\$350$ million daily. If workers strike and he loses $8\%$ for every hour of the strike, the initial amount is $a = 350,000,000$. The decay factor is calculated as $1 - 0.08 = 0.92$. The function is $f(x) = 350,000,000 \times 0.92^{x}$.

For a professional poker player starting with $\$5000$ who loses half his money every day, the initial amount is $a = 5000$. The decay factor is $0.5$. The function is $f(x) = 5000 \times 0.5^{x}$.

In the case of unicorns, the current population is $2000$, and it is increasing by $16\%$ every year. The initial amount is $a = 2000$, and the growth factor is $1 + 0.16 = 1.16$. The equation is $f(x) = 2000 \times 1.16^{x}$.

Round Rock has a population of $145,000$ and grows at a rate of $7\%$ annually. The initial amount is $a = 145,000$, and the growth factor is $1.07$. The equation is $f(x) = 145,000 \times 1.07^{x}$.

An investment of $\$350$ that doubles in value every year has an initial amount $a = 350$ and a growth factor $b = 2$. The equation is $f(x) = 350 \times 2^{x}$.

Under specific temperatures, $470$ bacterium of E. coli can triple every hour. The initial amount is $a = 470$ and the growth factor is $b = 3$. The equation is $f(x) = 470 \times 3^{x}$.

Del Valle has a population of $12,000$ and grows at a rate of $3.5\%$ per year. The initial amount is $a = 12,000$ and the factor is $1 + 0.035 = 1.035$. The equation is $f(x) = 12,000 \times 1.035^{x}$.

A savings account starts with $\$5150$ at an annual interest rate of $12.3\%$. The initial amount is $a = 5150$ and the growth factor is $1 + 0.123 = 1.123$. The equation is $f(x) = 5150 \times 1.123^{x}$.

An element with a mass of $730\text{ grams}$ decays by $25\%$ per minute. The initial amount is $a = 730$ and the decay factor is $1 - 0.25 = 0.75$. The equation is $f(x) = 730 \times 0.75^{x}$.

A mobile home purchased for $\$65,000$ depreciates at a rate of $8\%$ per year. The initial amount is $a = 65,000$ and the factor is $0.92$. The equation is $f(x) = 65,000 \times 0.92^{x}$.

A Honda Accord is purchased for $\$24,850$ and depreciates at $10\%$ per year. The initial amount is $a = 24,850$ and the decay factor is $0.90$. The equation is $f(x) = 24,850 \times 0.9^{x}$.

A mint condition Charizard Pokemon card is valued at $\$100,000$ and depreciates by $20\%$ every year. The initial amount is $a = 100,000$ and the factor is $0.80$. The equation is $f(x) = 100,000 \times 0.8^{x}$.

Procedural Solving of Exponential Equations

When solving equations where the variable is in the exponent, we often use logarithms to isolate the variable. The following problems require rounding answers to the nearest ten-thousandth (four decimal places).

For exercise 88, $9^{x} = 49$, applying the logarithm gives $x = \frac{\ln(49)}{\ln(9)} \approx 1.7712$. In exercise 89, $12^{x} = 13$, which results in $x = \frac{\ln(13)}{\ln(12)} \approx 1.0317$. Exercise 90, $5(18)^{x} = 26$, is solved by first dividing by $5$ to get $(18)^{x} = 5.2$, then finding $x = \frac{\ln(5.2)}{\ln(18)} \approx 0.5705$.

Exercise 91 states $3^{x - 5} = 5$. Taking the logarithm gives $x - 5 = \frac{\ln(5)}{\ln(3)}$, so $x = 5 + \frac{\ln(5)}{\ln(3)} \approx 6.4650$. Exercise 92 involves $3.4(3)^{-2x+2} - 9 = -4$. Adding $9$ gives $3.4(3)^{-2x+2} = 5$. Dividing by $3.4$ gives $3^{-2x+2} = 1.470588$. Taking logs: $-2x + 2 = \frac{\ln(1.470588)}{\ln(3)}$. Solving for $x$ yields approximately $0.8335$.

Exercise 93, $-6(5)^{8x+8} + 3 = -23$. Subtracting $3$ gives $-6(5)^{8x+8} = -26$, then dividing by $-6$ gives $5^{8x+8} = \frac{13}{3}$. This leads to $8x + 8 = \frac{\ln(\frac{13}{3})}{\ln(5)}$, resulting in $x \approx -0.8864$.

Exercise 94, $6(2)^{5x-6} - 4 = 50$. Adding $4$ gives $6(2)^{5x-6} = 54$, then $(2)^{5x-6} = 9$. Taking logs: $5x - 6 = \frac{\ln(9)}{\ln(2)}$, so $x \approx 1.8340$.

Exercise 95, $-2(3)^{7x+5} - 10 = -17$. Adding $10$ gives $-2(3)^{7x+5} = -7$, then $3^{7x+5} = 3.5$. Logging gives $7x + 5 = \frac{\ln(3.5)}{\ln(3)}$, so $x \approx -0.5471$.

Exercise 96, $-3(7)^{7x+9} + 6 = -6$. This results in $-3(7)^{7x+9} = -12$, then $7^{7x+9} = 4$. This leads to $7x + 9 = \frac{\ln(4)}{\ln(7)}$, resulting in $x \approx -1.1839$.

Exercise 97, $-3(9)^{x+1} + 6 = -58$. Subtracting $6$ gives $-3(9)^{x+1} = -64$, then $9^{x+1} = \frac{64}{3}$. Taking logs: $x + 1 = \frac{\ln(\frac{64}{3})}{\ln(9)}$, so $x \approx 0.3927$.

Systematic Resolution of Logarithmic Equations

Logarithmic equations require the application of logarithmic properties to condense terms when multiple logs exist on one side. Key properties include the product rule ($\log_{b}(m) + \log_{b}(n) = \log_{b}(m \times n)$), quotients rule ($\log_{b}(m) - \log_{b}(n) = \log_{b}(\frac{m}{n})$), and power rule ($p \times \log_{b}(m) = \log_{b}(m^{p})$).

In exercise 98, $\log_{3}(2x) = 3\log_{3}(6) - \log_{3}(4)$. We condense the right side to $\log_{3}(\frac{6^{3}}{4}) = \log_{3}(54)$. Setting arguments equal: $2x = 54 \rightarrow x = 27$. For 99, $\log_{4}(x) - 2\log_{4}(3) = \log_{4}(5)$. Condensing gives $\log_{4}(\frac{x}{3^{2}}) = \log_{4}(5)$, so $\frac{x}{9} = 5 \rightarrow x = 45$.

Exercise 100, $\log(5x) = \log(2x + 9)$. Setting arguments equal: $5x = 2x + 9 \rightarrow 3x = 9 \rightarrow x = 3$. Exercise 101, $\log(-4x + 7) = \log(3x)$, results in $-4x + 7 = 3x \rightarrow 7 = 7x \rightarrow x = 1$.

Exercise 102, $\log(4x) = \log(2) + 2\log(4)$. Condensing gives $\log(4x) = \log(2 \times 16) = \log(32)$, so $4x = 32 \rightarrow x = 8$. Exercise 103, $4\log(2) + \log(x) = \log(160)$. Condensing yields $\log(2^{4} \times x) = \log(160)$, so $16x = 160 \rightarrow x = 10$.

Exercise 104, $\log_{2}(-4x + 10) = \log_{2}(-3x + 10)$. This means $-4x + 10 = -3x + 10 \rightarrow -x = 0 \rightarrow x = 0$. Exercise 105, $\log_{6}(x) - 3\log_{6}(2) = \log_{6}(8)$. Condensing gives $\log_{6}(\frac{x}{2^{3}}) = \log_{6}(8)$, so $\frac{x}{8} = 8 \rightarrow x = 64$.

Exercise 106, $\log_{5}(2x) = \log_{5}(8) + 2\log_{5}(4)$. Condensing gives $\log_{5}(2x) = \log_{5}(8 \times 16) = \log_{5}(128)$, so $2x = 128 \rightarrow x = 64$.

For equations where the log is equal to a constant, rewrite in exponential form. Exercise 107, $\log_{4}(x) = 2$, becomes $x = 4^{2} = 16$. Exercise 108, $\log_{5}(4x - 3) = 0$, becomes $4x - 3 = 5^{0} = 1 \rightarrow 4x = 4 \rightarrow x = 1$. Exercise 109, $\log(3x - 8) = 1$. Since the base is $10$, this becomes $3x - 8 = 10^{1} \rightarrow 3x = 18 \rightarrow x = 6$.