About this video

• Video Title: Review: Solving Exponential and Log Equations
• Channel: Tom Kowalski
• Speakers: Tom Kowalski
• Duration: 17:12

Overview

This video provides a comprehensive review of techniques for solving exponential and logarithmic equations. It covers scenarios with the same base, different bases, and special cases, along with a detailed explanation of how to manipulate logarithmic properties and verify solutions.

Key takeaways

• Solving Exponential Equations with Same Base: If an equation can be written with two powers equal to each other and having the same base, set the exponents equal and solve.
• Handling Different Bases in Exponential Equations: When bases differ, rewrite all terms as powers of a common base (e.g., base 2). Use exponent properties for division and multiplication, then equate the exponents. Alternatively, take the logarithm of both sides to bring exponents down as multipliers.
• Logarithmic Equation Strategies:
  • If an equation is in the form log_b(f(x)) = log_b(g(x)), then f(x) = g(x).
  • Combine multiple log terms into a single log using properties of logarithms (product, quotient, power rules).
  • For log(expression) = constant, rewrite as an equivalent exponential equation.
• Domain Verification is Crucial: Always check solutions in the original logarithmic equation to ensure the input to the logarithm is positive. Solutions that result in a non-positive input are rejected.
• Logarithm Properties for Simplification: Use properties like log(a) + log(b) = log(ab), log(a) - log(b) = log(a/b), and n*log(a) = log(a^n) to simplify equations and make them easier to solve.

Here's a study guide organized to help you master solving exponential and logarithmic equations, based on the video content:

Study Guide: Solving Exponential and Logarithmic Equations

This guide breaks down the key techniques and considerations for solving exponential and logarithmic equations, as presented in the video.

---

I. Exponential Equations

A. Equations with the Same Base

• Principle: If you can express both sides of the equation as powers with the same base, you can equate the exponents.
  • Property Used: The one-to-one property of exponential functions. If $b^x = b^y$, then $x = y$.
• Steps:
  1. Identify if both sides of the equation can be written with the same base.
  2. Rewrite the terms using the common base.
  3. Set the exponents equal to each other.
  4. Solve the resulting (usually linear) equation.
• Example: $16^{x} / 1024 = 32^{x+1}$
  • Rewrite all bases as powers of 2:
    • $16 = 2^4$
    • $1024 = 2^{10}$
    • $32 = 2^5$
  • Substitute and apply exponent rules:
    • $(2^4)^x / 2^{10} = (2^5)^{x+1}$
    • $2^{4x} / 2^{10} = 2^{5(x+1)}$
    • $2^{4x - 10} = 2^{5x + 5}$
  • Equate exponents:
    • $4x - 10 = 5x + 5$
  • Solve for $x$:
    • $-15 = x$

B. Equations with Different Bases

• Challenge: Bases cannot be easily made the same.
• Method 1: Convert to a Common Base (if possible)
  • As shown in the example above, find a common base that can be used to express all terms.
• Method 2: Using Logarithms
  • Principle: Taking the logarithm of both sides allows you to use the power rule of logarithms ($log(a^b) = b \cdot log(a)$) to bring exponents down.
  • Steps:
    1. Isolate the exponential terms if necessary.
    2. Take the logarithm (base 10 or natural log, denoted as 'ln', are convenient for approximations) of both sides of the equation.
    3. Use the power rule to bring down the exponents.
    4. Solve the resulting equation, which will involve logarithmic terms. This often requires algebraic manipulation to isolate the variable.
  • Example: $5^{2m-3} = 2^{m+1}$
    • Take the natural log of both sides:
      • $ln(5^{2m-3}) = ln(2^{m+1})$
    • Apply the power rule:
      • $(2m - 3)ln(5) = (m + 1)ln(2)$
    • Distribute:
      • $2m \cdot ln(5) - 3 \cdot ln(5) = m \cdot ln(2) + ln(2)$
    • Gather terms with 'm' on one side and constants on the other:
      • $2m \cdot ln(5) - m \cdot ln(2) = ln(2) + 3 \cdot ln(5)$
    • Factor out 'm':
      • $m(2 \cdot ln(5) - ln(2)) = ln(2) + 3 \cdot ln(5)$
    • Solve for 'm':
      • $m = (ln(2) + 3 \cdot ln(5)) / (2 \cdot ln(5) - ln(2))$
    • Simplification for Calculator: Use log properties to combine terms for easier calculation:
      • $m = (ln(2) + ln(5^3)) / (ln(5^2) - ln(2))$
      • $m = ln(2 \cdot 125) / ln(25 / 2)$
      • $m = ln(250) / ln(12.5)$
    • Calculate the approximate value.

C. Exponential Equation Equaling a Constant

• Method: Rewrite the equation in logarithmic form.
• Steps:
  1. Isolate the exponential expression ($b^{\text{exponent}} = \text{constant}$).
  2. Rewrite in log form: $\text{exponent} = \log_b(\text{constant})$. If the base is 'e', use the natural logarithm.
  3. Solve the resulting equation.
• Example: $400 = 900 e^{-5t}$
  • Isolate the exponential term:
    • $400/900 = e^{-5t}$
    • $4/9 = e^{-5t}$
  • Rewrite in natural log form:
    • $ln(4/9) = -5t$
  • Solve for 't':
    • $t = ln(4/9) / -5$
  • Simplification for Calculator: Use log properties to remove the negative sign from the exponent if desired: $ln(4/9) = ln((9/4)^{-1}) = -ln(9/4)$.
    • $t = -ln(9/4) / -5 = ln(9/4) / 5$
  • Calculate the approximate value.

---

II. Logarithmic Equations

A. Logarithm Equaling a Logarithm (Same Base)

• Principle: If you have a single logarithm on each side of the equation with the same base, the inputs must be equal.
  • Property Used: The one-to-one property of logarithmic functions. If $log_b(f(x)) = log_b(g(x))$, then $f(x) = g(x)$.
• Steps:
  1. Use logarithm properties to combine terms on each side so that you have a single logarithm on the left and a single logarithm on the right, both with the same base. Ensure no multipliers are in front of the logs.
  2. Set the inputs (arguments) of the logarithms equal to each other.
  3. Solve the resulting equation.
  4. Crucially, verify solutions by plugging them back into the original equation to ensure the input to any logarithm is positive.
• Example: $log_2(x-1) + log_2(x-5) = log_2(3x+5)$
  • Combine logs on the left:
    • $log_2((x-1)(x-5)) = log_2(3x+5)$
  • Equate the inputs:
    • $(x-1)(x-5) = 3x+5$
  • Expand and rearrange into a quadratic equation:
    • $x^2 - 6x + 5 = 3x + 5$
    • $x^2 - 9x = 0$
  • Solve the quadratic (factoring):
    • $x(x-9) = 0$
    • Potential solutions: $x=0, x=9$
  • Verify Solutions:
    • For $x=0$: $log_2(0-1)$ is $log_2(-1)$, which is undefined. Reject $x=0$.
    • For $x=9$: $log_2(9-1) = log_2(8)$ (defined), $log_2(9-5) = log_2(4)$ (defined), $log_2(3(9)+5) = log_2(32)$ (defined). Check: $log_2(8) + log_2(4) = 3 + 2 = 5$. $log_2(32) = 5$. The equation holds. Accept $x=9$.

B. Logarithm Equaling a Constant

• Method: Rewrite the equation in exponential form.
• Steps:
  1. Use logarithm properties to combine terms on the side with the logarithm(s) into a single logarithm. Ensure no multipliers are in front of the log.
  2. Rewrite the equation in its equivalent exponential form: $\text{base}^{\text{constant}} = \text{input}$.
  3. Solve the resulting equation.
  4. Verify solutions to ensure the input to the logarithm in the original equation is positive.
• Example: $ln(x) - ln(x-1) = 2$
  • Combine logs on the left:
    • $ln(x / (x-1)) = 2$
  • Rewrite in exponential form (base is 'e'):
    • $e^2 = x / (x-1)$
  • Solve for x:
    • $e^2(x-1) = x$
    • $e^2x - e^2 = x$
    • $e^2x - x = e^2$
    • $x(e^2 - 1) = e^2$
    • $x = e^2 / (e^2 - 1)$
  • Verify Solution: Since $e \approx 2.718$, $e^2$ is greater than 1, so $e^2 - 1$ is positive. $e^2$ is positive. The result $x$ is positive.
    • Check $ln(x)$: $ln(e^2 / (e^2 - 1))$ is positive since $e^2 > e^2 - 1$.
    • Check $ln(x-1)$: $x - 1 = (e^2 / (e^2 - 1)) - 1 = (e^2 - (e^2 - 1)) / (e^2 - 1) = 1 / (e^2 - 1)$. This is also positive.
    • The solution is valid.

---

III. General Reminders

• Logarithm Domain: The argument (input) of a logarithm must always be positive ($> 0$).
• Logarithm Properties: Master these for simplifying expressions:
  • Product Rule: $log_b(M) + log_b(N) = log_b(MN)$
  • Quotient Rule: $log_b(M) - log_b(N) = log_b(M/N)$
  • Power Rule: $log_b(M^p) = p \cdot log_b(M)$
• Change of Base Formula: $log_b(a) = log_c(a) / log_c(b)$ (Useful for calculators or when bases are related, e.g., 8 and 2).
• One-to-One Functions: Both exponential ($b^x$) and logarithmic ($log_b(x)$) functions are one-to-one, meaning if their outputs are equal, their inputs must also be equal (under the appropriate conditions). This is the foundation for solving many of these equations.

---