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Zorluk: ZorAlgebraic Exponents and Radicals
If xx is a real number greater than 11 satisfying the exponential equation
(xx)x=x(xx)\left(x^x\right)^{\sqrt{x}} = x^{\left(x^{\sqrt{x}}\right)}
what is the value of xx?
  1. A
    32\frac{3}{2}
  2. B
    49\frac{4}{9}
  3. 94\frac{9}{4}Cevap
  4. D
    278\frac{27}{8}
  5. E
    8116\frac{81}{16}

Cevap

The value of xx is 94\frac{9}{4}.
Applying the exponent rule (ab)c=abc(a^b)^c = a^{bc} simplifies the left side to xxxx^{x\sqrt{x}}. Since the base x>1x > 1 is identical on both sides, we set the exponents equal: xx=xxx\sqrt{x} = x^{\sqrt{x}}. Expressing xxx\sqrt{x} as x3/2x^{3/2} yields x3/2=xxx^{3/2} = x^{\sqrt{x}}, which implies x=32\sqrt{x} = \frac{3}{2}. Squaring both sides gives x=94x = \frac{9}{4}.

Adım Adım Çözüm

1
Apply the power rule (ab)c=abc(a^b)^c = a^{bc} to the left-hand side of the equation.
(xx)x=xxx=xx3/2\left(x^x\right)^{\sqrt{x}} = x^{x \cdot \sqrt{x}} = x^{x^{3/2}}
Raising a power to another exponent requires multiplying the exponents: xx1/2=x1+1/2=x3/2x \cdot x^{1/2} = x^{1 + 1/2} = x^{3/2}.
2
Equate the exponents of the expressions on both sides, as the bases are equal and x>1x > 1.
xx=xxx \sqrt{x} = x^{\sqrt{x}}, which means x3/2=xxx^{3/2} = x^{\sqrt{x}}
If xA=xBx^A = x^B and x>1x > 1, then A=BA = B.
3
Equate exponents once more for the base xx.
32=x\frac{3}{2} = \sqrt{x}
Since the bases are identical (x>1x > 1), their exponents must be equal.
4
Square both sides to solve for xx.
x=(32)2=94x = \left(\frac{3}{2}\right)^2 = \frac{9}{4}
Squaring x\sqrt{x} isolates xx.

Anahtar Kavram

Properties of exponents and nested power rules with radicals
Tahmini Süre:2m 0s
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