Algebra 2 - Solving Exponential Equations & Logs

Solving Exponential Equations & Logs

Warmup

• Given $f(x) = 2^x$, determine the following:

  • a) $f(2) = 4$

  • b) $f(5) = 32$

  • c) $f(x) = 8$, then $x = 3$

  • d) $f(x) = 32$, then $x = 5$

Solving Exponential Equations

• Find the value of $x$ for the following problems (Use Desmos to confirm):

  • 1. $2^x = 16$ $x = 4$

  • 2. $3^x = 81$ $x = 4$

  • 3. $6^x = 36$ $x = 2$

  • 4. $5^x = 1$ $x = 0$

  • 5. $4^x = \frac{1}{4}$ $x = -1$

• Solving without Desmos:

  • 1. $2^x = 16$ $2^x = 2^4$, thus $x = 4$

  • 2. $3^x = 81$ $3^x = 3^4$, thus $x = 4$

  • 3. $6^x = 36$ $6^x = 6^2$, thus $x = 2$

  • 4. $5^x = 1$ $5^x = 5^0$, thus $x = 0$

  • 5. $4^x = \frac{1}{4}$ $4^x = 4^{-1}$, thus $x = -1$

• Solve for $x$: if $b^m = b^n$ implies $m = n$

  • 1. $2^{x+5} = 2^{4x+9}$

    • $x + 5 = 4x + 9$

    • $-3x = 4$

    • $x = -\frac{4}{3}$

  • 2. $7^{x+5} = 7^{8x-19}$

    • $x + 5 = 8x - 19$

    • $7x = 24$

    • $x = \frac{24}{7}$

  • 3. $8^{3x+5} = 8^{5x-3}$

    • $3x + 5 = 5x - 3$

    • $2x = 8$

    • $x = 4$

Multistep Equations

• Recall: SADMEP (Subtraction, Addition, Division, Multiplication, Exponents, Parentheses)

  • 1. $3 \cdot 2^x = 24$

    • $2^x = 8$

    • $2^x = 2^3$

    • $x = 3$

  • 2. $2^{x+1} = 16$

    • $2^{x+1} = 2^4$

    • $x + 1 = 4$

    • $x = 3$

  • 3. $6^x \div 2 = 18$

    • $6^x = 36$

    • $6^x = 6^2$

    • $x = 2$

  • 4. $.34125^x = .34122^{x+9}$

    • This appears to be an error in the original document.

Investigation Activity

• 1) Enter the equation $y = 2^x$ into Desmos

• 2) Use the graph to fill in the table and sketch the graph.

• 3) Characteristics of the graph of $y = 2^x$

  • Domain: $(-\infty, \infty)$

  • Range: $(0, \infty)$

  • x-intercept(s): none

  • y-intercept: $(0, 1)$

  • Equation of an asymptote: $y = 0$

• 4) Create a table of values for the INVERSE of $y = 2^x$.

• This inverse is represented by $y = log_2(x)$

• 5) Enter $y = log_2 x$ into Desmos (using the calculator -> functions -> loga).

• 6) Characteristics of the graph of $y = log_2 x$

  • Domain: $(0, \infty)$

  • Range: $(-\infty, \infty)$

  • x-intercept(s): $(1, 0)$

  • y-intercept: none

  • Equation of an asymptote: $x = 0$

• 7) Enter in the equation $y = x$

  • Exponential growth and logs of the same base are inverses of each other. Visually, that means they are a reflection over the $y = x$ line.

  • MANTRA: Answer to a log question is always an exponent.

• 8) Examine the following logarithmic functions:

  • A. $y = log_2(x-3)$

    • Vertical Asymptote: $x = 3$

    • x-intercept: $(4, 0)$

    • Domain: $(3, \infty)$

    • Range: $(-\infty, \infty)$

  • B. $y = log_2(x+7)$

    • Vertical Asymptote: $x = -7$

    • x-intercept: $(-6, 0)$

    • Domain: $(-7, \infty)$

    • y-intercept: $(0, log_2 7)$

    • Range: $(-\infty, \infty)$

• 9) Create a table of values for the INVERSE of $y = 10^x$

• This inverse is represented by $y = log_{10} x$

• 10) Create a table of values for the INVERSE of $y = e^x$

• This inverse is represented by $y = ln(x)$

Comparing Exponential and Logarithmic Graphs

• Graphs of $y = 2^x$ and $y = log_2 x$

• Discuss: what do you notice about the graphs? What are the similarities and differences?

  • Similarities: They are inverses of each other and reflected over the line $y = x$.

  • Differences: Their domains, ranges, intercepts, and asymptotes are different.

• Use both graphs to fill in the following blanks:

  • The domain of $y = 2^x$ is equal to the range of $y = log_2 x$.

  • The range of $y = 2^x$ is equal to the domain of $y = log_2 x$.

  • The x-intercept of $y = 2^x$ is related to the y-intercept of $y = log_2 x$.

  • The y-intercept of $y = 2^x$ is related to the x-intercept of $y = log_2 x$.

• What is the relationship of the asymptotes on the graph of $y = 2^x$ versus $y = log_2 x$?

  • The horizontal asymptote of $y = 2^x$ becomes the vertical asymptote of $y = log_2 x$.

• 10) Using the table of values from the equations $y = 2^x$ and $y = log_2 x$

• Class Discussion: How does the solution of a logarithm connect to an exponential equation?

  • The solution to a logarithm is the exponent to which the base must be raised to obtain the argument of the logarithm.

Evaluating a Logarithm

• Determining what exponent is needed for base a to be raised to equal the value of b: $log_a b$

• Evaluate the logarithms:

  • $log_a a = 1$

  • $log_b b = 1$

  • a. $log_3 3 = 1$

  • b. $log_3 9 = 2$

  • c. $log_4 4 = 1$

  • d. $log_4 64 = 3$

  • e. $log_5 1 = 0$

  • f. $log_5 5 = 1$

  • g. $log_5 25 = 2$

  • h. $log_9 81 = 2$

  • i. $log_7 7 = 1$

  • j. $log_4 16 = 2$

• There are two special logs:

  • Natural log: $ln \Rightarrow log_e = ln(x)$

  • Common log: $log \Rightarrow log_{10} (x)$

• Evaluate the logarithms:

  • a. $log 100 = log_{10} 10^2 = 2$

  • b. $log 10 = 1$

  • c. $ln e = 1$

  • d. $ln 1 = 0$

• Think back to Part 1 when we used Desmos to solve for $x$ in the equation: $2^x = 5$

• We can now use logarithms to solve as another method for solving for an exponent!

• Rewrite the exponential equation as a logarithm using Desmos functions -> loga.

• Solve the exponential equation by converting to logarithms.

  • 1. $3^x = 4$ $x = log_3 4$

  • 2. $3^x = 12$ $x = log_3 12$

  • 3. $4^x = 100$ $x = log_4 100$

  • 4. $4^x = 52$ $x = log_4 52$

  • 5. $1.34^x = 7$ $x = log_{1.34} 7$

  • 6. $1.34^x = 5$ $x = log_{1.34} 5$

  • 7. $0.97^x = 30$ $x = log_{0.97} 30$

  • 8. $5^x = 0$ $x = log_5 0$

• Discuss: Why do you think we got the answer we did for number 8?

  • Logarithms are not defined for non-positive numbers (0 and negative numbers), because you cannot raise a positive base to any power and get a non-positive result.

Tying Logarithms Back to Exponential Growth and Decay

• 1. The equation $c(t) = 40,000 (1.065)^t$ represents the cost of tuition, in dollars, as a function of $t$, the number of years since 2020. Approximately when will the tuition cost $60,000 ?

  • Method one:

    • $y = 40,000 (1.065)^t$

    • $y = 60,000$

  • Method two:

    • $60000 = 40000 (1.065)^t$

    • $1.5 = (1.065)^t$

    • $t = log_{1.065} 1.5$

• Approximately when will the tuition cost $60,000? 6.43 years.

• 2. Brad is doing a science experiment. The equation represents the amount of bacteria over time: $f(x) = 82(1.257)^x$ where $x$ is the number of days. Determine $f(x) = 650$.

  • $650 = 82 (1.257)^x$

  • $\frac{650}{82} = (1.257)^x$

  • $x = log_{1.257} (\frac{650}{82})$ $\approx 9.05 \text{ days}$

• 3. A small business bought a snowplow for $30,000 in 2015. The plow depreciates (loses value) by 33% every year after its purchase. The equation that represents the value of the snowplow over time is: $y = 30,000 (.67)^x$ where $x$ represents the number of years since purchase. After how many years will the plow be worth $9,000? Round to the nearest year.

  • $9000 = 30000 (.67)^x$

  • $(\frac{9000}{30000}) = (.67)^x$

  • $0.3 = (0.67)^x$

  • $x = log_{0.67} 0.3$

  • $x ≈ 3$

  • 3 years

Applications: Solving logarithmic equations using the same base idea.

• Solve the equations for $x$.

  • 1. $log<em>2 x = log</em>2(3x-10)$

    • $x = 3x - 10$

    • $-2x = -10$

    • $x = 5$

  • 2. $log<em>3 (5x+1) = log</em>3 (2x+25)$

    • $5x + 1 = 2x + 25$

    • $3x = 24$

    • $x = 8$

  • 3. $ln(4x) = ln 12$

    • $4x = 12$

    • $x = 3$