Question

Difficulty: HardExponential Functions and Equations

If 16x1104x2+1=016^{x-1} - 10 \cdot 4^{x-2} + 1 = 0, what is the sum of all values of xx that satisfy the equation?

Answer: 2

Answer

The sum of all values of xx that satisfy the equation is 2.
By writing 16x116^{x-1} as (4x)216\frac{(4^x)^2}{16} and 104x210 \cdot 4^{x-2} as 104x16\frac{10 \cdot 4^x}{16}, multiplying the entire equation by 16 yields (4x)210(4x)+16=0(4^x)^2 - 10(4^x) + 16 = 0. Factoring this quadratic equation gives (4x2)(4x8)=0(4^x - 2)(4^x - 8) = 0, which yields 4x=24^x = 2 and 4x=84^x = 8. Solving these exponential equations results in x=0.5x = 0.5 and x=1.5x = 1.5. Adding these values together gives 0.5+1.5=20.5 + 1.5 = 2.

Step-by-Step Solution

1
Express each exponential term in the equation using a base of 4.
42x16104x16+1=0\frac{4^{2x}}{16} - \frac{10 \cdot 4^x}{16} + 1 = 0
Since 16=4216 = 4^2, we can write 16x1=(42)x1=42x2=42x1616^{x-1} = (4^2)^{x-1} = 4^{2x-2} = \frac{4^{2x}}{16}. Similarly, 4x2=4x164^{x-2} = \frac{4^x}{16}.
2
Multiply every term in the equation by 16 to eliminate the fractions.
42x104x+16=04^{2x} - 10 \cdot 4^x + 16 = 0
This simplifies the expression into a standard polynomial form that is quadratic in terms of 4x4^x.
3
Use the substitution u=4xu = 4^x (where u>0u > 0) to transform the equation into a quadratic equation in terms of uu, and solve it.
u210u+16=0    (u2)(u8)=0    u=2u^2 - 10u + 16 = 0 \implies (u - 2)(u - 8) = 0 \implies u = 2 or u=8u = 8
Substituting uu allows us to factor the quadratic expression using standard factoring techniques.
4
Substitute 4x4^x back for uu and solve for xx using base 2 conversion.
For u=2u = 2: 4x=2    22x=21    x=0.54^x = 2 \implies 2^{2x} = 2^1 \implies x = 0.5. For u=8u = 8: 4x=8    22x=23    x=1.54^x = 8 \implies 2^{2x} = 2^3 \implies x = 1.5.
We must convert the solutions for the intermediate variable uu back to the original variable xx.
5
Calculate the sum of all found values of xx.
0.5+1.5=20.5 + 1.5 = 2
The question asks for the sum of all real values of xx that satisfy the equation.

Key Concept

Solving exponential equations of quadratic form by using variable substitution and exponent laws.
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