What is the answer to 2^{x+1}-2^{x-1}-8=0 ?

The answer to 2^{x+1}-2^{x-1}-8=0 is x=(ln(16/3))/(ln(2))

Solve for x: 2^{x+1}-2^{x-1}-8=0

The solution is x=(ln(16/3))/(ln(2))

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2^{x + 1} - 2^{x - 1} - 8 = 0

x = \frac{ln ( \frac{16}{3} )}{ln ( 2 )}

Solution steps

2^{x + 1} - 2^{x - 1} - 8 = 0

Factor

2^{x + 1} - 2^{x - 1} - 8 : \frac{1}{2} (3 \cdot 2^{x} - 16)

\frac{1}{2} (3 \cdot 2^{x} - 16) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

3 \cdot 2^{x} - 16 = 0

Solve

3 \cdot 2^{x} - 16 = 0 : x = \frac{ln ( \frac{16}{3} )}{ln ( 2 )}

x = \frac{ln ( \frac{16}{3} )}{ln ( 2 )}

3^{5 x - 2} = 6

4^{5 - 9 x} = \frac{1}{8 ^{x - 2}}

\frac{1}{2} = 4^{x}

225 e^{\frac{x}{4}} = 45

e^{x - 3} = 6

What is the answer to 2^{x+1}-2^{x-1}-8=0 ?
The answer to 2^{x+1}-2^{x-1}-8=0 is x=(ln(16/3))/(ln(2))
Solve for x: 2^{x+1}-2^{x-1}-8=0
The solution is x=(ln(16/3))/(ln(2))