Math Problem Statement

Log2(x+5)=2+log2(x+2)

Solution

We are solving the equation:

log2(x+5)=2+log2(x+2)\log_2(x + 5) = 2 + \log_2(x + 2)

Step 1: Simplify using logarithmic properties

We know the logarithmic property: logb(a)logb(c)=logb(ac)\log_b(a) - \log_b(c) = \log_b\left(\frac{a}{c}\right) Rewriting the equation, we isolate the logarithmic terms: log2(x+5)log2(x+2)=2\log_2(x + 5) - \log_2(x + 2) = 2

Now apply the property: log2(x+5x+2)=2\log_2\left(\frac{x + 5}{x + 2}\right) = 2

Step 2: Eliminate the logarithm

To eliminate the logarithm, rewrite the equation in exponential form: x+5x+2=22\frac{x + 5}{x + 2} = 2^2 x+5x+2=4\frac{x + 5}{x + 2} = 4

Step 3: Solve for xx

Multiply both sides by x+2x + 2 (ensuring x+20x + 2 \neq 0): x+5=4(x+2)x + 5 = 4(x + 2) Expand the right-hand side: x+5=4x+8x + 5 = 4x + 8 Simplify: 58=4xx5 - 8 = 4x - x 3=3x-3 = 3x x=1x = -1

Step 4: Verify the solution

Substitute x=1x = -1 into the original equation to ensure it satisfies the domain of the logarithms:

  • For log2(x+5)\log_2(x + 5), we have log2(1+5)=log2(4)\log_2(-1 + 5) = \log_2(4), which is valid.
  • For log2(x+2)\log_2(x + 2), we have log2(1+2)=log2(1)\log_2(-1 + 2) = \log_2(1), which is valid.

Now substitute into the original equation: log2(1+5)=2+log2(1+2)\log_2(-1 + 5) = 2 + \log_2(-1 + 2) log2(4)=2+log2(1)\log_2(4) = 2 + \log_2(1) 2=22 = 2

The solution is verified.

Final Answer:

x=1x = -1

Do you have any questions or want a detailed explanation of any step?


  1. What are logarithmic properties, and how are they applied?
  2. How do you handle domains when solving logarithmic equations?
  3. Why is verifying solutions important in logarithmic equations?
  4. What happens if a solution is not in the domain of the logarithm?
  5. Can you explain how to rewrite logarithmic equations in exponential form?

Tip: Always check the domain of logarithmic functions to ensure solutions are valid.

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Math Problem Analysis

Mathematical Concepts

Logarithms
Properties of Logarithms
Exponential Equations

Formulas

log_b(a) - log_b(c) = log_b(a/c)
log_b(a) = c => a = b^c

Theorems

Properties of Logarithms
Exponential-Logarithmic Relationship

Suitable Grade Level

Grades 10-12