What is the answer to 2^x-12*2^x+32=0 ?

The answer to 2^x-12*2^x+32=0 is x=(ln(32/11))/(ln(2))

Solve for x: 2^x-12*2^x+32=0

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

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2^{x} - 12 \cdot 2^{x} + 32 = 0

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

Solution steps

2^{x} - 12 \cdot 2^{x} + 32 = 0

Add similar elements:

2^{x} - 12 \cdot 2^{x} = - 11 \cdot 2^{x}

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

Move

32

to the right side

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

Divide both sides by

- 11

2^{x} = \frac{32}{11}

Apply exponent rules

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

Solve

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

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

15.7 (0.98)^{x} = 11 (0.99)^{x}

4^{x} (5^{2} x) = 10

40000 = 5400 \cdot 1.03 5^{x}

e^{9 x} = 4

2^{3 x} - 2^{7 x - 2} = 0

What is the answer to 2^x-12*2^x+32=0 ?
The answer to 2^x-12*2^x+32=0 is x=(ln(32/11))/(ln(2))
Solve for x: 2^x-12*2^x+32=0
The solution is x=(ln(32/11))/(ln(2))