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Question

Difficulty: MediumExponential Functions and Equations

If 92x−1=27x+239^{2x - 1} = \frac{27^{x + 2}}{3}, what is the value of xx?

  1. A
    3
  2. 7Answer
  3. C
    8
  4. D
    9

Answer

The correct value of xx is 7.
To solve the equation 92x−1=27x+239^{2x - 1} = \frac{27^{x + 2}}{3}, write all bases as powers of 3: 9=329 = 3^2 and 27=3327 = 3^3. Substituting these into the equation gives (32)2x−1=(33)x+231(3^2)^{2x - 1} = \frac{(3^3)^{x + 2}}{3^1}. Applying the power rule of exponents, we get 34x−2=33x+6313^{4x - 2} = \frac{3^{3x + 6}}{3^1}. Using the quotient rule of exponents on the right side, we subtract the exponent in the denominator (which is 1) from the exponent in the numerator: 33x+6−1=33x+53^{3x + 6 - 1} = 3^{3x + 5}. Now we have 34x−2=33x+53^{4x - 2} = 3^{3x + 5}. Since the bases are the same, we set the exponents equal to each other: 4x−2=3x+54x - 2 = 3x + 5. Solving for xx by subtracting 3x3x and adding 2 to both sides gives the correct value of 7.

Step-by-Step Solution

1
Express all bases as powers of 3.
(32)2x−1=(33)x+231(3^2)^{2x - 1} = \frac{(3^3)^{x + 2}}{3^1}
To solve exponential equations with different bases, we rewrite them using a common base to apply exponent laws.
2
Apply the power of a power rule to simplify exponents.
34x−2=33x+6313^{4x - 2} = \frac{3^{3x + 6}}{3^1}
Multiply exponents when raising a power to another power: (am)n=am⋅n(a^m)^n = a^{m \cdot n}.
3
Apply the quotient rule to simplify the right side of the equation.
34x−2=33x+53^{4x - 2} = 3^{3x + 5}
Subtract the exponent of the denominator from the numerator when dividing: aman=am−n\frac{a^m}{a^n} = a^{m-n}.
4
Set the exponents equal to each other and solve the linear equation.
x=7x = 7
Since the bases on both sides are equal, their exponents must be equal: 4x−2=3x+54x - 2 = 3x + 5.

Key Concept

Solving exponential equations by expressing terms with a common base and applying exponent properties.
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Topics

Exponential Functions and Equations

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