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Difficulty: HardLogarithmic and Exponential Expressions and Equations

For all positive real numbers xx, which of the following expressions is equivalent to log⁡3(27x4)−log⁡3(3x2)\log_3(27x^4) - \log_3(3x^2)?

  1. A
    2+6log⁡3(x)2 + 6\log_3(x)
  2. B
    log⁡3(24x2)\log_3(24x^2)
  3. 2+2log⁡3(x)2 + 2\log_3(x)Answer
  4. D
    4log⁡3(x)4\log_3(x)
  5. E
    3+2log⁡3(x)3 + 2\log_3(x)

Answer

2+2log⁡3(x)2 + 2\log_3(x)
The correct answer is found by applying the quotient property of logarithms to combine the terms, yielding log⁡3(9x2)\log_3(9x^2). Then, applying the product property splits this into log⁡3(9)+log⁡3(x2)\log_3(9) + \log_3(x^2). Finally, evaluating log⁡3(9)=2\log_3(9) = 2 and using the power property to rewrite log⁡3(x2)\log_3(x^2) as 2log⁡3(x)2\log_3(x) yields the simplified expression 2+2log⁡3(x)2 + 2\log_3(x).

Step-by-Step Solution

1
Apply the quotient property of logarithms: log⁡b(A)−log⁡b(B)=log⁡b(AB)\log_b(A) - \log_b(B) = \log_b\left(\frac{A}{B}\right).
log⁡3(27x43x2)\log_3\left(\frac{27x^4}{3x^2}\right)
To combine the two logarithmic terms into a single logarithm.
2
Simplify the algebraic expression inside the logarithm.
log⁡3(9x2)\log_3(9x^2)
Dividing the coefficients (27÷3=927 \div 3 = 9) and subtracting the exponents of the variable xx (4−2=24 - 2 = 2).
3
Apply the product property of logarithms: log⁡b(CD)=log⁡b(C)+log⁡b(D)\log_b(CD) = \log_b(C) + \log_b(D).
log⁡3(9)+log⁡3(x2)\log_3(9) + \log_3(x^2)
To separate the constant and variable parts of the logarithmic argument.
4
Evaluate the numerical logarithm and apply the power property of logarithms: log⁡b(yk)=klog⁡b(y)\log_b(y^k) = k\log_b(y).
2+2log⁡3(x)2 + 2\log_3(x)
Since 32=93^2 = 9, log⁡3(9)=2\log_3(9) = 2, and the exponent of xx can be moved in front of the logarithm as a multiplier.

Key Concept

Logarithmic and Exponential Expressions and Equations

Alternative Method

Alternatively, expand each logarithm first using the product and power properties of logarithms:
1. log⁡3(27x4)=log⁡3(27)+log⁡3(x4)=3+4log⁡3(x)\log_3(27x^4) = \log_3(27) + \log_3(x^4) = 3 + 4\log_3(x)
2. log⁡3(3x2)=log⁡3(3)+log⁡3(x2)=1+2log⁡3(x)\log_3(3x^2) = \log_3(3) + \log_3(x^2) = 1 + 2\log_3(x)
Subtracting the second expanded expression from the first yields:
(3+4log⁡3(x))−(1+2log⁡3(x))=3−1+4log⁡3(x)−2log⁡3(x)=2+2log⁡3(x)(3 + 4\log_3(x)) - (1 + 2\log_3(x)) = 3 - 1 + 4\log_3(x) - 2\log_3(x) = 2 + 2\log_3(x).
Estimated Time:1m 30s
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Topics

Logarithmic and Exponential Expressions and Equations

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