What is the answer to 2^x+2^{x+3}+2^{x-1}= 19/4 ?

The answer to 2^x+2^{x+3}+2^{x-1}= 19/4 is x=-1

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2^x+2^{x+3}+2^{x-1}= 19/4

2^{x} + 2^{x + 3} + 2^{x - 1} = \frac{19}{4}

x = - 1

Solution steps

2^{x} + 2^{x + 3} + 2^{x - 1} = \frac{19}{4}

Factor

2^{x} + 2^{x + 3} + 2^{x - 1} : 19 \cdot 2^{x - 1}

19 \cdot 2^{x - 1} = \frac{19}{4}

Divide both sides by

19

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

Apply exponent rules

x - 1 = - 2

Solve

x - 1 = - 2 : x = - 1

x = - 1

2 \times 9^{x} = 162

7^{x + 5} = 4

9^{x} - 4 \times 3^{x} + 3 = 0

\frac{1000}{1 + 39 e ^{- 0.42 x}} = 630

3 = e^{2 x}

What is the answer to 2^x+2^{x+3}+2^{x-1}= 19/4 ?
The answer to 2^x+2^{x+3}+2^{x-1}= 19/4 is x=-1
Solve for x: 2^x+2^{x+3}+2^{x-1}= 19/4
The solution is x=-1