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Zorluk: KolaySurds and Indices

If 2x1+2x+1=3202^{x-1} + 2^{x+1} = 320, what is the value of xx?

  1. A
    55
  2. B
    66
  3. 77Cevap
  4. D
    88

Cevap

77
Factoring 2x12^{x-1} from the left side gives 2x1(1+22)=52x1=3202^{x-1}(1 + 2^2) = 5 \cdot 2^{x-1} = 320. Dividing both sides by 55 yields 2x1=64=262^{x-1} = 64 = 2^6. Equating the powers gives x1=6x - 1 = 6, which solves to x=7x = 7.

Adım Adım Çözüm

1
Factor out the common exponential term from the left-hand side
2x1(1+22)=320    2x1(1+4)=320    52x1=3202^{x-1}(1 + 2^2) = 320 \implies 2^{x-1}(1 + 4) = 320 \implies 5 \cdot 2^{x-1} = 320
Applying the distributive law of exponents to simplify the sum.
2
Divide both sides by 5
2x1=3205=642^{x-1} = \frac{320}{5} = 64
Isolate the exponential term with base 2.
3
Express 64 as a power of 2 and equate the exponents
64=26    x1=6    x=764 = 2^6 \implies x - 1 = 6 \implies x = 7
Since the bases are equal (22), the exponents must be equal.

Anahtar Kavram

Laws of Indices and Exponential Equations
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