What is the answer to n=((2^{n+4}-2^{n+3}))/((2^{n+4))} ?

The answer to n=((2^{n+4}-2^{n+3}))/((2^{n+4))} is n= 1/2

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n=((2^{n+4}-2^{n+3}))/((2^{n+4))}

n = \frac{( 2 ^{n + 4} - 2 ^{n + 3} )}{( 2 ^{n + 4} )}

n = \frac{1}{2}

Decimal Notation

n = 0.5

Solution steps

n = \frac{( 2 ^{n + 4} - 2 ^{n + 3} )}{( 2 ^{n + 4} )}

Apply exponent rules

n = \frac{1}{2}

Verify Solutions:

n = \frac{1}{2}

True

The solution is

n = \frac{1}{2}

\frac{8 1 ^{2}}{( 72 9 ^{1} ) ^{- x}} = (9^{2})^{x}

2^{x - 4} = 4. 2^{x - 6}

2 = 1.0 6^{x}

x - e^{- x} = 0

x, 2^{x + 4 y} = 1

What is the answer to n=((2^{n+4}-2^{n+3}))/((2^{n+4))} ?
The answer to n=((2^{n+4}-2^{n+3}))/((2^{n+4))} is n= 1/2
Solve for n: n=((2^{n+4}-2^{n+3}))/((2^{n+4))}
The solution is n= 1/2