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Logarithms - Solving Logarithmic Equations: The Search for 'x'-advanced

Grade 9CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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To solve logarithmic equations, we primarily use the relationship between logs and exponents: log⁡b(x)=y\log_b(x) = y is equivalent to by=xb^y = x, where b>0,b≠1b > 0, b \neq 1, and x>0x > 0.

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Product Rule: The sum of two logarithms with the same base can be written as the logarithm of the product of their arguments: log⁡b(M)+log⁡b(N)=log⁡b(MN)\log_b(M) + \log_b(N) = \log_b(MN).

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Quotient Rule: The difference of two logarithms with the same base can be written as the logarithm of the quotient of their arguments: log⁡b(M)−log⁡b(N)=log⁡b(MN)\log_b(M) - \log_b(N) = \log_b\left(\frac{M}{N}\right).

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Power Rule: A coefficient in front of a logarithm can be moved to the exponent of the argument: k⋅log⁡b(M)=log⁡b(Mk)k \cdot \log_b(M) = \log_b(M^k).

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One-to-One Property: If log⁡b(M)=log⁡b(N)\log_b(M) = \log_b(N), then M=NM = N. This is frequently used to remove logs from both sides of an equation.

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Change of Base Formula: This is vital for solving equations where bases are different: log⁡b(a)=log⁡k(a)log⁡k(b)\log_b(a) = \frac{\log_k(a)}{\log_k(b)}.

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Extraneous Solutions: Always check the final values of xx in the original equation. The argument of any logarithm must be greater than zero (x>0x > 0).

📐Formulae

log⁡b(x)=y  ⟺  by=x\log_b(x) = y \iff b^y = x

log⁡b(MN)=log⁡b(M)+log⁡b(N)\log_b(MN) = \log_b(M) + \log_b(N)

log⁡b(MN)=log⁡b(M)−log⁡b(N)\log_b\left(\frac{M}{N}\right) = \log_b(M) - \log_b(N)

log⁡b(Mk)=klog⁡b(M)\log_b(M^k) = k \log_b(M)

log⁡b(a)=log⁡k(a)log⁡k(b)\log_b(a) = \frac{\log_k(a)}{\log_k(b)}

log⁡a(a)=1 and log⁡a(1)=0\log_a(a) = 1 \text{ and } \log_a(1) = 0

log⁡bn(am)=mnlog⁡b(a)\log_{b^n}(a^m) = \frac{m}{n} \log_b(a)

💡Examples

Problem 1:

Solve for xx: log⁡2(x)+log⁡2(x−2)=3\log_2(x) + \log_2(x - 2) = 3.

Solution:

  1. Use the Product Rule to combine the logs: log⁡2(x(x−2))=3\log_2(x(x - 2)) = 3.
  2. Convert the logarithmic equation into exponential form: x(x−2)=23x(x - 2) = 2^3.
  3. Simplify the equation: x2−2x=8x^2 - 2x = 8.
  4. Form a quadratic equation: x2−2x−8=0x^2 - 2x - 8 = 0.
  5. Factor the quadratic: (x−4)(x+2)=0(x - 4)(x + 2) = 0.
  6. This gives two potential solutions: x=4x = 4 or x=−2x = -2.
  7. Check for extraneous solutions: If x=−2x = -2, the term log⁡2(x)\log_2(x) becomes log⁡2(−2)\log_2(-2), which is undefined. If x=4x = 4, both log⁡2(4)\log_2(4) and log⁡2(4−2)\log_2(4 - 2) are defined.
  8. Therefore, x=4x = 4.

Explanation:

In this problem, we combined terms using log properties, solved the resulting quadratic equation, and discarded the solution that made the log argument negative.

Problem 2:

Solve for xx: log⁡x(3)+log⁡3(x)=2.5\log_x(3) + \log_3(x) = 2.5.

Solution:

  1. Use the Change of Base formula or the property log⁡x(3)=1log⁡3(x)\log_x(3) = \frac{1}{\log_3(x)}.
  2. Let y=log⁡3(x)y = \log_3(x). The equation becomes: 1y+y=2.5\frac{1}{y} + y = 2.5.
  3. Multiply by yy to clear the fraction: 1+y2=2.5y1 + y^2 = 2.5y.
  4. Rewrite as a standard quadratic: y2−2.5y+1=0y^2 - 2.5y + 1 = 0.
  5. Multiply by 22 to remove decimals: 2y2−5y+2=02y^2 - 5y + 2 = 0.
  6. Factor the quadratic: (2y−1)(y−2)=0(2y - 1)(y - 2) = 0.
  7. Solve for yy: y=12y = \frac{1}{2} or y=2y = 2.
  8. Substitute back y=log⁡3(x)y = \log_3(x): Case 1: log⁡3(x)=12  ⟹  x=312=3\log_3(x) = \frac{1}{2} \implies x = 3^{\frac{1}{2}} = \sqrt{3}. Case 2: log⁡3(x)=2  ⟹  x=32=9\log_3(x) = 2 \implies x = 3^2 = 9.

Explanation:

This advanced problem uses a substitution method (y=log⁡3(x)y = \log_3(x)) after identifying that log⁡x(3)\log_x(3) and log⁡3(x)\log_3(x) are reciprocals.

Problem 3:

Solve for xx: 2log⁡(x)=log⁡(2)+log⁡(3x−4)2\log(x) = \log(2) + \log(3x - 4).

Solution:

  1. Use the Power Rule on the left side: log⁡(x2)=log⁡(2)+log⁡(3x−4)\log(x^2) = \log(2) + \log(3x - 4).
  2. Use the Product Rule on the right side: log⁡(x2)=log⁡(2(3x−4))\log(x^2) = \log(2(3x - 4)).
  3. Since the logs are equal, their arguments must be equal (One-to-One Property): x2=2(3x−4)x^2 = 2(3x - 4).
  4. Expand and rearrange: x2=6x−8  ⟹  x2−6x+8=0x^2 = 6x - 8 \implies x^2 - 6x + 8 = 0.
  5. Factor the quadratic: (x−4)(x−2)=0(x - 4)(x - 2) = 0.
  6. Potential solutions: x=4x = 4 or x=2x = 2.
  7. Check: For x=4x=4, 3(4)−4=8>03(4)-4=8 > 0. For x=2x=2, 3(2)−4=2>03(2)-4=2 > 0. Both are valid.

Explanation:

We used the power rule to bring the coefficient inside the log as an exponent, then equated the arguments once both sides had a single log term.