Logarithmic Form Solutions

Grade 11 · algebra · 100 practice problems · read aloud

🔊 Listen to this explanation

Logarithmic Form Solutions

🔍 What is it and Why is it Useful?

A logarithm answers the question: "To what exponent must we raise a base to get a certain number?" It's the inverse of exponentiation. This is incredibly useful for solving equations where the variable is in the exponent, which is common in real-world scenarios like calculating compound interest, pH levels, and earthquake intensity (Richter scale).

📝 Step-by-Step Guide

To solve an exponential equation using logarithms:

  1. Isolate the exponential expression on one side of the equation.
  2. Apply the logarithm (common log, log₁₀, or natural log, ln) to both sides of the equation.
  3. Use the Power Rule of logarithms to bring the exponent down in front: log(bˣ) = x · log(b).
  4. Solve for the variable algebraically.
  5. Check your solution by substituting it back into the original equation.

✨ Visual Examples

Example 1: Solve 5ˣ = 25

  1. The exponential expression, 5ˣ, is already isolated.
  2. Take the log of both sides: log(5ˣ) = log(25)
  3. Apply the Power Rule: x · log(5) = log(25)
  4. Solve for x: x = log(25) / log(5)
  5. Since 25 is 5², we know log(25) = 2 · log(5). So, x = (2 · log(5)) / log(5) = 2.

Example 2: Solve 3e²ˣ = 60

  1. Isolate e²ˣ: e²ˣ = 20
  2. Take the natural log (ln) of both sides: ln(e²ˣ) = ln(20)
  3. Apply the Power Rule and simplify ln(e) to 1: 2x · ln(e) = 2x · 1 = ln(20)
  4. Solve for x: x = ln(20) / 2 ≈ 1.497

⚠️ Common Mistakes

  • Misapplying Log Rules: log(a + b) is NOT log(a) + log(b). The correct rule is for multiplication: log(a·b) = log(a) + log(b).
  • Forgetting the Base: When you see "log" with no base, it is base 10. "ln" is base e.
  • Incorrect Isolation: Always isolate the exponential term before taking the logarithm of both sides.

💡 Tips & Tricks

  • Memory Aid: Remember that "A logarithm is an exponent." It's in the definition!
  • Shortcut: If you have an equation like aˣ = b and 'b' can be written as a power of 'a', you can solve it by inspection (like in Example 1) without needing a calculator.
  • Strategy: If the base is e, always use the natural logarithm (ln). If the base is 10, use the common log (log). For other bases, you can use either.

🎯 Practice Suggestions

To master logarithmic solutions:

  1. Start by practicing with simple equations where the answer is an integer.
  2. Move on to problems requiring a calculator, rounding your answers to 3 decimal places.
  3. Create a "mistake journal" to track and learn from the common errors listed above.
  4. Try writing your own exponential word problems and then solve them using logarithms.

Practice problems

6 of the 100, worked through step by step — try them before opening the answer.

1 log₃(27) = ?

Hint: Consider what power the base must be raised to in order to equal the argument. For example, log₂(8) asks: 2 to what power equals 8?

Show the answer

Answer: 3

  1. Understand the meaning of log₃(27). It means: "To what power must we raise 3 to get 27?" In other words, if log₃(27) = x, then 3^x = 27.
  2. Express 27 as a power of 3. We know 3^1 = 3, 3^2 = 9, 3^3 = 27. So 27 = 3^3.
  3. Substitute 27 with 3^3 in the equation from
  4. 3^x = 3^3.
  5. Since the bases are the same (base 3), the exponents must be equal. Therefore, x = 3.
  6. Conclusion. log₃(27) = 3. Final answer: 3

We are asked to evaluate log₃(27).

2 log₄(64) = ?

Hint: Consider what power you need to raise the base to in order to get the argument. For example, log₂(8) asks '2 to what power equals 8?'

Show the answer

Answer: 3

  1. We need to find the exponent x such that 4^x = 64
  2. Recognize that 64 can be written as 4^3 because 4 × 4 × 4 = 16 × 4 = 64
  3. Therefore, 4^3 = 64
  4. This means log₄(64) = 3

The answer is 3.

3 log₃(x) + log₃(x+6) = 3

Hint: Combine the logarithmic terms using the product rule, then convert to exponential form to solve for the variable.

Show the answer

Answer: 3

  1. Apply the product rule for logarithms: log₃(x) + log₃(x+6) = log₃(x(x+6)) = 3
  2. Convert to exponential form: x(x+6) = 3³ = 27
  3. Expand and rearrange: x² + 6x - 27 = 0
  4. Factor the quadratic: (x+9)(x-3) = 0
  5. Solve for x: x = -9 or x = 3
  6. Check domain restrictions: log₃(x) requires x > 0, so x = -9 is extraneous The valid solution is x = 3.

4 7^x = 49. Express solution using logarithms

Hint: Remember that logarithms are the inverse operation of exponents. For example, if 3^y = 9, then y = log₃(9).

Show the answer

Answer: 2

  1. Start with the equation 7^x = 49
  2. Recognize that 49 can be written as 7^2, since 7 × 7 = 49
  3. Substitute 7^2 for 49: 7^x = 7^2
  4. Since the bases are equal, the exponents must be equal: x = 2
  5. Express in logarithmic form: x = log₇(49)

The answer is 2.

5 8^x = 64. Express solution using logarithms

Hint: Remember that logarithmic form is the inverse of exponential form. For example, if a^b = c, then b = log_a(c).

Show the answer

Answer: x = log_8(64)

  1. Start with the exponential equation: 8^x = 64
  2. To express this in logarithmic form, we write: x = log_8(64)
  3. This means x is the exponent we need to raise 8 to in order to get 64
  4. The solution in logarithmic form is x = log_8(64)

6 7^x = 50. Express solution using logarithms

Hint: To solve exponential equations, convert to logarithmic form using the definition: if a^b = c, then b = logₐ(c)

Show the answer

Answer: x = log₇(50)

  1. Start with the equation 7^x = 50
  2. Apply the definition of logarithms: if a^b = c, then b = logₐ(c)
  3. Here, a = 7, b = x, and c = 50
  4. Therefore, x = log₇(50) The solution in logarithmic form is x = log₇(50)
Practise this topic — 10 free problems, no signup →