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Zorluk: OrtaExponents, Radicals, and Algebraic Expressions

If 4x+4x+4x+4x=2104^x + 4^x + 4^x + 4^x = 2^{10}, what is the value of xx?

  1. A
    54\frac{5}{4}
  2. B
    22
  3. C
    52\frac{5}{2}
  4. 44Cevap
  5. E
    55

Cevap

The value of xx is 44.
Combining the four identical terms on the left side gives 44x=41+x4 \cdot 4^x = 4^{1+x}. Converting 44 to base 22 yields (22)x+1=22x+2(2^2)^{x+1} = 2^{2x+2}. Setting this equal to the right side 2102^{10} gives the linear equation 2x+2=102x + 2 = 10, which solves to x=4x = 4.

Adım Adım Çözüm

1
Combine the repeated addition on the left side of the equation.
4x+4x+4x+4x=44x=4x+14^x + 4^x + 4^x + 4^x = 4 \cdot 4^x = 4^{x+1}
Adding four identical terms is equivalent to multiplying the term by 4.
2
Convert all bases to 2 so that both sides can be compared.
4x+1=(22)x+1=22(x+1)=22x+24^{x+1} = (2^2)^{x+1} = 2^{2(x+1)} = 2^{2x+2}
Since 4=224 = 2^2, applying the power of a power rule (am)n=amn(a^m)^n = a^{mn} expresses the left side with base 2.
3
Equate the exponents and solve for xx.
2x+2=10    2x=8    x=42x + 2 = 10 \implies 2x = 8 \implies x = 4
If bu=bvb^u = b^v for a positive base b1b \neq 1, then u=vu = v.

Anahtar Kavram

Combining like exponential terms by factoring and applying base conversion rules aman=am+na^{m} \cdot a^{n} = a^{m+n} and (am)n=amn(a^m)^n = a^{mn}.
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