What is n given that 2*2n+1=[2*2*2]15

Break down left side: Now, let's look at the left side of the equation. 2*2^(n+1) can be written as 2^(1)*2^(n+1) = 2^(n+1+1) = 2^(n+2)Set exponents equal: We can now set the exponents equal to each other since the bases are the same. So, n+2 = 45Subtract to solve for n: Subtract 2 from both sides to solve for n. n = 45 - 2 n = 43

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What is $n$ given that $2^{2n+1}=[2\cdot2\cdot2]^{15}$

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Q. What is

n

given that

2^{2n+1}=[2\cdot2\cdot2]^{15}

Break down left side: Now, let's look at the left side of the equation. $\newline$ $2\times 2^{(n+1)}$ can be written as $2^{1}\times 2^{(n+1)} = 2^{(n+1+1)} = 2^{(n+2)}$
Set exponents equal: We can now set the exponents equal to each other since the bases are the same. $\newline$ So, $n+2 = 45$
Subtract to solve for n: Subtract $2$ from both sides to solve for $n$ . $\newline$ $n = 45 - 2$ $\newline$ $n = 43$