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Zorluk: OrtaLogarithmic and Exponential Expressions and Equations

If 92x−1=27x+49^{2x - 1} = 27^{x + 4}, what is the value of xx?

  1. A
    5
  2. B
    -13
  3. 14Cevap
  4. D
    6
  5. E
    -4

Cevap

14
Converting both bases to 3 yields (32)2x−1=(33)x+4(3^2)^{2x - 1} = (3^3)^{x + 4}. Applying the exponent power rule gives 34x−2=33x+123^{4x - 2} = 3^{3x + 12}. Equating the exponents results in 4x−2=3x+124x - 2 = 3x + 12. Subtracting 3x3x from both sides gives x−2=12x - 2 = 12, and adding 2 to both sides results in x=14x = 14. This matches the correct value of 14.

Adım Adım Çözüm

1
Express both sides of the equation with a common base of 3.
(32)2x−1=(33)x+4(3^2)^{2x - 1} = (3^3)^{x + 4}
Since 9=329 = 3^2 and 27=3327 = 3^3, rewriting the bases allows us to equate the exponents later.
2
Apply the power of a power property, (am)n=amn(a^m)^n = a^{mn}, to simplify the exponents on both sides.
32(2x−1)=33(x+4)3^{2(2x - 1)} = 3^{3(x + 4)} which simplifies to 34x−2=33x+123^{4x - 2} = 3^{3x + 12}
To simplify an exponent raised to another power, multiply the exponents, ensuring the multiplier is distributed to both terms inside each exponent expression.
3
Set the exponents equal to each other and solve the resulting linear equation for xx.
4x−2=3x+12  ⟹  x=144x - 2 = 3x + 12 \implies x = 14
If two exponential expressions with the same positive base (other than 1) are equal, their exponents must be equal.

Anahtar Kavram

Solving exponential equations by expressing bases with a common base and equating the exponents.
Tahmini Süre:1m 30s
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Konular

Logarithmic and Exponential Expressions and Equations

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