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Zorluk: OrtaSurds and Indices

Consider the exponential equation:

4x32x+2+32=04^x - 3 \cdot 2^{x+2} + 32 = 0

Determine the sum of all real values of xx that satisfy this equation.

Cevap: 5

Cevap

5
By applying the laws of indices, the original expression transforms into a quadratic equation in terms of 2x2^x. Solving y212y+32=0y^2 - 12y + 32 = 0 yields y=4y=4 and y=8y=8, which correspond precisely to x=2x=2 and x=3x=3. Their sum is 5.

Adım Adım Çözüm

1
Express all terms with a common base of 2.
4x4^x becomes (2x)2(2^x)^2 and 2x+22^{x+2} becomes 42x4 \cdot 2^x.
Creating a common base allows the equation to be transformed into a standard polynomial form.
2
Rewrite the equation using the new terms.
(2x)212(2x)+32=0(2^x)^2 - 12(2^x) + 32 = 0
Simplifying the coefficients makes it easier to spot the quadratic structure.
3
Perform a substitution to solve the quadratic equation.
Letting y=2xy = 2^x gives y212y+32=0y^2 - 12y + 32 = 0. Factoring yields (y4)(y8)=0(y - 4)(y - 8) = 0, so y=4y = 4 or y=8y = 8.
Substitution converts a complex exponential equation into a simple quadratic one.
4
Solve for the original variable xx.
2x=4    x=22^x = 4 \implies x = 2, and 2x=8    x=32^x = 8 \implies x = 3.
The question asks for the values of xx, not the intermediate substitution variable yy.
5
Calculate the sum of all valid xx values.
2+3=52 + 3 = 5
This addresses the specific final requirement of the question stem.

Anahtar Kavram

Solving exponential equations reducible to quadratics using index laws.
Tahmini Süre:1m 30s
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