<span>If we are to compare which among the given fractions is smaller, we first determine the decimal equivalent of each. If we are to solve for those, we determine that 3/4 is equal to 0.75 and 1/4 is equal to 0.25. This is done by dividing the numerator by the denominator. Comparing 0.75 and 0.25, we can conclude that 0.75 or 3/4 is greater compared to 0.25 or 1/4. </span>
Answer:
x = 1 or x = -1
Step-by-step explanation:
Given equation:
![x^{100}-4^x \cdot x^{98}-x^2+4^x=0](https://tex.z-dn.net/?f=x%5E%7B100%7D-4%5Ex%20%5Ccdot%20x%5E%7B98%7D-x%5E2%2B4%5Ex%3D0)
Factor out -1:
![\implies -1(-x^{100}+4^x \cdot x^{98}+x^2-4^x)=0](https://tex.z-dn.net/?f=%5Cimplies%20-1%28-x%5E%7B100%7D%2B4%5Ex%20%5Ccdot%20x%5E%7B98%7D%2Bx%5E2-4%5Ex%29%3D0)
Divide both sides by -1:
![\implies -x^{100}+4^x \cdot x^{98}+x^2-4^x=0](https://tex.z-dn.net/?f=%5Cimplies%20-x%5E%7B100%7D%2B4%5Ex%20%5Ccdot%20x%5E%7B98%7D%2Bx%5E2-4%5Ex%3D0)
Rearrange the terms:
![\implies 4^x \cdot x^{98}-4^x-x^{100}+x^2=0](https://tex.z-dn.net/?f=%5Cimplies%204%5Ex%20%5Ccdot%20x%5E%7B98%7D-4%5Ex-x%5E%7B100%7D%2Bx%5E2%3D0)
![\textsf{Apply exponent rule} \quad a^{b+c}=a^b \cdot a^c \quad \textsf{to }x^{100}:](https://tex.z-dn.net/?f=%5Ctextsf%7BApply%20exponent%20rule%7D%20%5Cquad%20a%5E%7Bb%2Bc%7D%3Da%5Eb%20%5Ccdot%20a%5Ec%20%5Cquad%20%5Ctextsf%7Bto%20%7Dx%5E%7B100%7D%3A)
![\implies 4^x \cdot x^{98}-4^x-x^{98}x^2+x^2=0](https://tex.z-dn.net/?f=%5Cimplies%204%5Ex%20%5Ccdot%20x%5E%7B98%7D-4%5Ex-x%5E%7B98%7Dx%5E2%2Bx%5E2%3D0)
Factor the first two terms and the last two terms separately:
![\implies 4^x(x^{98}-1)-x^2(x^{98}-1)=0](https://tex.z-dn.net/?f=%5Cimplies%204%5Ex%28x%5E%7B98%7D-1%29-x%5E2%28x%5E%7B98%7D-1%29%3D0)
Factor out the common term
:
![\implies (4^x-x^2)(x^{98}-1)=0](https://tex.z-dn.net/?f=%5Cimplies%20%284%5Ex-x%5E2%29%28x%5E%7B98%7D-1%29%3D0)
<u>Zero Product Property</u>: If a ⋅ b = 0 then either a = 0 or b = 0 (or both).
Using the <u>Zero Product Property</u>, set each factor equal to zero and solve for x (if possible):
![\begin{aligned}x^{98}-1 & = 0 & \quad \textsf{or} \quad \quad4^x-x^2 & = 0 \\x^{98} & =1 & 4^x & = x^2 \\x & = 1, -1 & \textsf{no}& \textsf{ solutions for } x \in \mathbb{R}\end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7Dx%5E%7B98%7D-1%20%26%20%3D%200%20%26%20%5Cquad%20%5Ctextsf%7Bor%7D%20%5Cquad%20%5Cquad4%5Ex-x%5E2%20%26%20%3D%200%20%5C%5Cx%5E%7B98%7D%20%26%20%3D1%20%26%204%5Ex%20%26%20%3D%20x%5E2%20%5C%5Cx%20%26%20%3D%201%2C%20-1%20%26%20%5Ctextsf%7Bno%7D%26%20%5Ctextsf%7B%20solutions%20for%20%7D%20x%20%5Cin%20%5Cmathbb%7BR%7D%5Cend%7Baligned%7D)
Therefore, the solutions to the given equation are: x = 1 or x = -1
Learn more here:
brainly.com/question/27751281
brainly.com/question/21186424
Answer:
-5
Step-by-step explanation:
f(2) = 2 * f(1) + 1
f(2) = 2 * -3 + 1
f(2) = -6 + 1
f(2) = -5