What is the answer to ((2^{n+1}+2^n))/((2^{n+4)-2^{n-1})}=1 ?

The answer to ((2^{n+1}+2^n))/((2^{n+4)-2^{n-1})}=1 is No Solution for n\in\mathbb{R}

Solve for n: ((2^{n+1}+2^n))/((2^{n+4)-2^{n-1})}=1

The solution is No Solution for n\in\mathbb{R}

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\frac{( 2 ^{n + 1} + 2 ^{n} )}{( 2 ^{n + 4} - 2 ^{n - 1} )} = 1

No Solution for n \in R

Solution steps

\frac{( 2 ^{n + 1} + 2 ^{n} )}{( 2 ^{n + 4} - 2 ^{n - 1} )} = 1

Multiply both sides by

2^{n + 4} - 2^{n - 1}

\frac{2 ^{n + 1} + 2 ^{n}}{2 ^{n + 4} - 2 ^{n - 1}} (2^{n + 4} - 2^{n - 1}) = 1 \cdot (2^{n + 4} - 2^{n - 1})

Simplify

2^{n + 1} + 2^{n} = 2^{n + 4} - 2^{n - 1}

Subtract

2^{n + 4} - 2^{n - 1}

from both sides

2^{n + 1} + 2^{n} - (2^{n + 4} - 2^{n - 1}) = 2^{n + 4} - 2^{n - 1} - (2^{n + 4} - 2^{n - 1})

Simplify

2^{n + 1} + 2^{n} - 2^{n + 4} + 2^{n - 1} = 0

Factor

2^{n + 1} + 2^{n} - 2^{n + 4} + 2^{n - 1} : - 25 \cdot 2^{n - 1}

- 25 \cdot 2^{n - 1} = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

2^{n - 1} = 0

Solve

2^{n - 1} = 0 :

No Solution for

n \in R

No Solution for n \in R

8^{x - 3} = 4^{x + 1}

(- 3)^{x} = - 243

4^{x} - 2^{x + 4} = 80

3^{5} \cdot 3^{2} \cdot 0. 3^{6} = 3^{x}

2^{x} = 256

What is the answer to ((2^{n+1}+2^n))/((2^{n+4)-2^{n-1})}=1 ?
The answer to ((2^{n+1}+2^n))/((2^{n+4)-2^{n-1})}=1 is No Solution for n\in\mathbb{R}
Solve for n: ((2^{n+1}+2^n))/((2^{n+4)-2^{n-1})}=1
The solution is No Solution for n\in\mathbb{R}