What is the answer to 4^x+1+4^x-1=34 ?

The answer to 4^x+1+4^x-1=34 is x=(ln(17))/(2ln(2))

Solve for x: 4^x+1+4^x-1=34

The solution is x=(ln(17))/(2ln(2))

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4^{x} + 1 + 4^{x} - 1 = 34

x = \frac{ln ( 17 )}{2 ln ( 2 )}

Decimal Notation

x = 2.04373 \dots

Solution steps

4^{x} + 1 + 4^{x} - 1 = 34

Group like terms

4^{x} + 4^{x} + 1 - 1 = 34

Add similar elements:

4^{x} + 4^{x} = 2 \cdot 4^{x}

2 \cdot 4^{x} + 1 - 1 = 34

1 - 1 = 0

2 \cdot 4^{x} = 34

Divide both sides by

2

4^{x} = 17

Apply exponent rules

x ln (4) = ln (17)

Solve

x ln (4) = ln (17) : x = \frac{ln ( 17 )}{2 ln ( 2 )}

x = \frac{ln ( 17 )}{2 ln ( 2 )}

5^{x - 1} = \frac{1}{625}

1.56 = 1. 5^{x}

4^{3 x} = 4^{2 (x + 7)}

2^{2 x + 2} = 128

5^{x + 1} = 255

What is the answer to 4^x+1+4^x-1=34 ?
The answer to 4^x+1+4^x-1=34 is x=(ln(17))/(2ln(2))
Solve for x: 4^x+1+4^x-1=34
The solution is x=(ln(17))/(2ln(2))