MHB Solving Exponential Equations: Need Help Understanding 2 Questions

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To solve the exponential equation (2^x + 2)^2, the correct expansion is 2^(2x) + 2^(x+2) + 4, which differs from the initial incorrect answer of 22x + 2x + 4. The middle term was identified as the source of confusion. For the second question, the simplification of 84x/32x to (84/32)x is correct, and it can be further simplified to 27x by reducing the fraction. Understanding these steps clarifies the solutions to both questions.
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I need help with understanding two questions:
1. (2x+2)2
The answer I got was: 22x+2x++4, but according to the answers that is wrong. Can any explain why?

2. 84x/32x
I simplified this to become (84/32)x, but I don't know how to go from there... The answer is apparently 27x, but I don't understand how that can be...

Would appreciate the help enormously,
//APRIL
 
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Nevermind, I understood question 2 now, but question 1 is still a mystery to me...
 
linapril said:
I need help with understanding two questions:
1. (2x+2)2
The answer I got was: 22x+2x++4, but according to the answers that is wrong. Can any explain why?

I think the middle term is wrong. The first and the last terms look good though. Check your work and if you still don't see it then post your attempt and we'll help you sort it out :)
 
linapril said:
I need help with understanding two questions:
1. (2x+2)2
The answer I got was: 22x+2x++4, but according to the answers that is wrong. Can any explain why?

2. 84x/32x
I simplified this to become (84/32)x, but I don't know how to go from there... The answer is apparently 27x, but I don't understand how that can be...

Would appreciate the help enormously,
//APRIL

$\displaystyle \begin{align*} \left( 2^x + 2 \right)^2 &= \left( 2^x \right)^2 + 2\cdot 2\cdot 2^x + 2^2 \\ &= 2^{2x} + 2^{x + 2} + 2^2 \end{align*}$
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...

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