{:[(2^(x)+4^(y)+8^(z)=328)/(" \# ")],[x","y","z=??],[(x","y","z)in+z]:}

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Rewrite Equation: First, let's rewrite the equation in a more familiar form, using the fact that 4 = 2^2 and 8 = 2^3. This will allow us to express all terms with the same base, which is 2. 2^x + (2^2)^y + (2^3)^z = 328 Now apply the power of a power rule, which states that (a^b)^c = a^(b*c). 2^x + 2^(2y) + 2^(3z) = 328Apply Power Rule: Next, we need to find integer values of x, y, and z that satisfy the equation. Since 328 is not a power of 2, we know that x, y, and z must be such that their powers of 2 add up to 328. We can start by looking for powers of 2 that are close to 328 and see if we can find a combination that works.Find Values: We know that 2^8 = 256, which is the largest power of 2 less than 328. So, let's try z = 1, which makes the term 2^(3z) = 2^3 = 8. 2^x + 2^(2y) + 8 = 328 Subtract 8 from both sides to isolate the terms with x and y. 2^x + 2^(2y) = 320Try z = 1: Now, we look for powers of 2 that add up to 320. We know that 2^6 = 64, and 5 * 64 = 320. So, let's try y = 3, which makes the term 2^(2y) = 2^6 = 64. 2^x + 64 = 320 Subtract 64 from both sides to find the value of 2^x. 2^x = 256Subtract 8: Since we know that 2^8 = 256, we can conclude that x = 8. 2^8 = 256Look for Powers: Now we have found values for x, y, and z that satisfy the equation: x = 8, y = 3, z = 1 Let's check if these values satisfy the original equation: 2^x + 4^y + 8^z = 2^8 + 4^3 + 8^1 = 256 + 64 + 8 = 328

$\begin{array}{c}\frac{2^{x}+4^{y}+8^{z}=328}{\text { \# }} \\ x, y, z=? ? \\ (x, y, z) \in+z\end{array}$

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2x+4y+8z=328 # x,y,z=??(x,y,z)∈+z \begin{array}{c}\frac{2^{x}+4^{y}+8^{z}=328}{\text { \# }} \\ x, y, z=? ? \\ (x, y, z) \in+z\end{array} # 2x+4y+8z=328​x,y,z=??(x,y,z)∈+z​

$\begin{array}{c}\frac{2^{x}+4^{y}+8^{z}=328}{\text { \# }} \\ x, y, z=? ? \\ (x, y, z) \in+z\end{array}$