What is the answer to 2^{x-2}-3*2^x+8=0 ?

The answer to 2^{x-2}-3*2^x+8=0 is x=(ln(32/11))/(ln(2))

Solve for x: 2^{x-2}-3*2^x+8=0

The solution is x=(ln(32/11))/(ln(2))

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2^{x - 2} - 3 \cdot 2^{x} + 8 = 0

x = \frac{ln ( \frac{32}{11} )}{ln ( 2 )}

Decimal Notation

x = 1.54056 \dots

Solution steps

2^{x - 2} - 3 \cdot 2^{x} + 8 = 0

Factor

2^{x - 2} - 3 \cdot 2^{x} + 8 : \frac{1}{4} (- 11 \cdot 2^{x} + 32)

\frac{1}{4} (- 11 \cdot 2^{x} + 32) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

- 11 \cdot 2^{x} + 32 = 0

Solve

- 11 \cdot 2^{x} + 32 = 0 : x = \frac{ln ( \frac{32}{11} )}{ln ( 2 )}

x = \frac{ln ( \frac{32}{11} )}{ln ( 2 )}

2^{x^{2} + 7 x - 38} = 6 4^{x - 6}

79 = 3 3^{x}

15 = 20 + 30 e^{- \frac{t}{20}}

9^{x} = 3^{x} + 12

2000 = 8^{x}

What is the answer to 2^{x-2}-3*2^x+8=0 ?
The answer to 2^{x-2}-3*2^x+8=0 is x=(ln(32/11))/(ln(2))
Solve for x: 2^{x-2}-3*2^x+8=0
The solution is x=(ln(32/11))/(ln(2))