What is the answer to 2^{x+1}+2^x-2^x-2=88 ?

The answer to 2^{x+1}+2^x-2^x-2=88 is x=(ln(45))/(ln(2))

Solve for x: 2^{x+1}+2^x-2^x-2=88

The solution is x=(ln(45))/(ln(2))

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2^{x + 1} + 2^{x} - 2^{x} - 2 = 88

x = \frac{ln ( 45 )}{ln ( 2 )}

Decimal Notation

x = 5.49185 \dots

Solution steps

2^{x + 1} + 2^{x} - 2^{x} - 2 = 88

Factor

2^{x + 1} + 2^{x} - 2^{x} - 2 : 2 (2^{x} - 1)

2 (2^{x} - 1) = 88

Divide both sides by

2

2^{x} - 1 = 44

Move

1

to the right side

2^{x} = 45

Apply exponent rules

x ln (2) = ln (45)

Solve

x ln (2) = ln (45) : x = \frac{ln ( 45 )}{ln ( 2 )}

x = \frac{ln ( 45 )}{ln ( 2 )}

2 = (1.05)^{n}

2^{x - 2} = 8 x

e^{c} = 2

2 7^{x} = (\frac{1}{3})^{- 3}

3^{4 x + 4} + 3^{4 x} = 3^{5}

What is the answer to 2^{x+1}+2^x-2^x-2=88 ?
The answer to 2^{x+1}+2^x-2^x-2=88 is x=(ln(45))/(ln(2))
Solve for x: 2^{x+1}+2^x-2^x-2=88
The solution is x=(ln(45))/(ln(2))