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Zorluk: OrtaAlgebraic Exponents and Radicals

If xx is a real number satisfying the exponential equation 9x+132x+1=1629^{x+1} - 3^{2x+1} = 162, what is the value of 4x4^x?

  1. A
    4
  2. B
    6
  3. 8Cevap
  4. D
    16
  5. E
    64

Cevap

8
By converting 9x+19^{x+1} to 32x+23^{2x+2} and factoring out 32x3^{2x}, the equation simplifies to 632x=1626 \cdot 3^{2x} = 162. Dividing by 6 gives 32x=273^{2x} = 27, so 2x=32x = 3 and x=32x = \frac{3}{2}. Raising 4 to the power of 32\frac{3}{2} yields (4)3=8(\sqrt{4})^3 = 8.

Adım Adım Çözüm

1
Rewrite terms with a common base of 3.
9x+1=(32)x+1=32(x+1)=32x+29^{x+1} = (3^2)^{x+1} = 3^{2(x+1)} = 3^{2x+2}
Since 9 is a power of 3 (9=329 = 3^2), applying exponent rules converts the equation to base 3.
2
Factor out the common exponential expression 32x3^{2x}.
32x+232x+1=32x3232x31=32x(93)=632x3^{2x+2} - 3^{2x+1} = 3^{2x} \cdot 3^2 - 3^{2x} \cdot 3^1 = 3^{2x}(9 - 3) = 6 \cdot 3^{2x}
Using product rule of exponents (3a+b=3a3b3^{a+b} = 3^a \cdot 3^b) allows combining like terms.
3
Solve for xx.
632x=162    32x=27    32x=33    2x=3    x=326 \cdot 3^{2x} = 162 \implies 3^{2x} = 27 \implies 3^{2x} = 3^3 \implies 2x = 3 \implies x = \frac{3}{2}
Dividing both sides by 6 yields 32x=273^{2x} = 27, and equating exponents of matching bases gives x=32x = \frac{3}{2}.
4
Evaluate the target expression 4x4^x.
43/2=(41/2)3=23=84^{3/2} = (4^{1/2})^3 = 2^3 = 8
Substituting x=32x = \frac{3}{2} into 4x4^x means taking the square root of 4 and raising it to the third power.

Anahtar Kavram

Solving exponential equations using common bases and exponent properties
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