What is the answer to 4^x+2^{1+2x}=20 ?

The answer to 4^x+2^{1+2x}=20 is x=(ln(20/3))/(2ln(2))

Solve for x: 4^x+2^{1+2x}=20

The solution is x=(ln(20/3))/(2ln(2))

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4^{x} + 2^{1 + 2 x} = 20

x = \frac{ln ( \frac{20}{3} )}{2 ln ( 2 )}

Decimal Notation

x = 1.36848 \dots

Solution steps

4^{x} + 2^{1 + 2 x} = 20

Apply exponent rules

2 x ln (2) = ln (\frac{20}{3})

Solve

2 x ln (2) = ln (\frac{20}{3}) : x = \frac{ln ( \frac{20}{3} )}{2 ln ( 2 )}

x = \frac{ln ( \frac{20}{3} )}{2 ln ( 2 )}

1 + x = 2^{x}

86400 = 50000 (1 + \frac{1}{5})^{x}

x^{l o g_{3} (x) + 1} = 9

3^{(x - 1)} + 3^{x} = 4

0. 5^{x} = \frac{1}{16}

What is the answer to 4^x+2^{1+2x}=20 ?
The answer to 4^x+2^{1+2x}=20 is x=(ln(20/3))/(2ln(2))
Solve for x: 4^x+2^{1+2x}=20
The solution is x=(ln(20/3))/(2ln(2))