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what is nn given that 22n+1=[2⋅2⋅2]152^{2n+1}=[2\cdot2\cdot2]^{15}

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Full solution

Q. what is nn given that 22n+1=[2⋅2⋅2]152^{2n+1}=[2\cdot2\cdot2]^{15}
  1. Break down left side: Now, let's look at the left side of the equation.\newline2×2(n+1)2\times 2^{(n+1)} can be written as 21×2(n+1)=2(n+1+1)=2(n+2)2^{1}\times 2^{(n+1)} = 2^{(n+1+1)} = 2^{(n+2)}
  2. Set exponents equal: We can now set the exponents equal to each other since the bases are the same.\newlineSo, n+2=45n+2 = 45
  3. Subtract to solve for n: Subtract 22 from both sides to solve for nn.\newlinen=45−2n = 45 - 2\newlinen=43n = 43

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