Math Problem Statement

2log3(x+7)=log39+2

Solution

We are tasked with solving the equation:

2log3(x+7)=log39+22\log_3{(x+7)} = \log_3{9} + 2

Step 1: Use logarithmic properties

We know that logb(ac)=clogba\log_b{(a^c)} = c\log_b{a}, so we can start by simplifying the left-hand side:

2log3(x+7)=log3(x+7)22\log_3{(x+7)} = \log_3{(x+7)^2}

Thus, the equation becomes:

log3(x+7)2=log39+2\log_3{(x+7)^2} = \log_3{9} + 2

Step 2: Simplify log39\log_3{9}

We know that 9=329 = 3^2, so:

log39=log332=2\log_3{9} = \log_3{3^2} = 2

Thus, the equation simplifies to:

log3(x+7)2=2+2\log_3{(x+7)^2} = 2 + 2

log3(x+7)2=4\log_3{(x+7)^2} = 4

Step 3: Remove the logarithm by exponentiation

Since logby=x\log_b{y} = x is equivalent to bx=yb^x = y, we raise both sides of the equation to powers of 3:

(x+7)2=34(x+7)^2 = 3^4

(x+7)2=81(x+7)^2 = 81

Step 4: Solve for xx

Take the square root of both sides:

x+7=±81x+7 = \pm \sqrt{81}

x+7=±9x+7 = \pm 9

Thus, we have two cases:

  1. x+7=9x+7 = 9x=2x = 2
  2. x+7=9x+7 = -9x=16x = -16

Step 5: Check the solutions

For logarithmic equations, the argument of the logarithm must be positive. Let's check both solutions:

  • For x=2x = 2, x+7=9x + 7 = 9, which is positive.
  • For x=16x = -16, x+7=9x + 7 = -9, which is negative and not allowed.

Thus, the only valid solution is:

x=2\boxed{x = 2}


Would you like more details or have any questions?

Here are 5 related questions:

  1. What are the properties of logarithms used in solving logarithmic equations?
  2. How do you handle logarithmic equations when the base is not explicitly given?
  3. What are the conditions for the arguments of logarithmic functions to be valid?
  4. How would the solution change if the equation was in base 10 instead of base 3?
  5. How do you verify solutions of logarithmic equations?

Tip: Always check the domain of the logarithmic function to ensure the argument is positive before accepting solutions.

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Math Problem Analysis

Mathematical Concepts

Logarithms
Logarithmic Equations
Algebra

Formulas

log_b(a^c) = c log_b(a)
log_b(y) = x ⟹ b^x = y

Theorems

Logarithmic Identity
Exponentiation Rule for Logarithms

Suitable Grade Level

Grades 10-12