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why is $2^{log_2(n-1) + 1} = 2(n-1)?$ I tried using the formula $b^{log_b(n)} = log_b(b^n) = n$ $\,$ but can't seem to get $2(n-1)$ please help

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  • $\begingroup$ Hint: $2^{x+1}=2\times 2^x$ $\endgroup$ – gammatester Oct 30 '18 at 17:01
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Notice $$ 2^{\log_2(n-1)+1}=2^{\log_2(n-1)}2^1=(n-1)2^1=2(n-1) $$ Here I used the fact that $2^{a+1}=2^a\cdot2$ (or more generally, $2^{a+b}=2^a2^b$)

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The $1$ isn’t part of the logarithm argument. Use the following property of exponents.

$$a^{b+c} = a^b\cdot a^c$$

$$\implies 2^{\log_2 (n-1)+1} = 2^{\log_2 (n-1)} \cdot 2^1 \implies 2^{\log_2 (n-1)} \cdot 2$$

Now, you can use the property of logarithms that you mentioned.

$$a^{\log_a b} = b$$

$$\implies 2^{\log_2 (n-1)} \cdot 2 = 2(n-1)$$

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