What is the answer to 10^x+10^{x+1}=22 ?

The answer to 10^x+10^{x+1}=22 is x=(ln(2))/(ln(10))

Solve for x: 10^x+10^{x+1}=22

The solution is x=(ln(2))/(ln(10))

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1 0^{x} + 1 0^{x + 1} = 22

x = \frac{ln ( 2 )}{ln ( 10 )}

Decimal Notation

x = 0.30102 \dots

Solution steps

1 0^{x} + 1 0^{x + 1} = 22

Factor

1 0^{x} + 1 0^{x + 1} : 11 \cdot 1 0^{x}

11 \cdot 1 0^{x} = 22

Divide both sides by

11

1 0^{x} = 2

Apply exponent rules

x ln (10) = ln (2)

Solve

x ln (10) = ln (2) : x = \frac{ln ( 2 )}{ln ( 10 )}

x = \frac{ln ( 2 )}{ln ( 10 )}

e^{x} = x^{3}

\frac{16 e ^{x}}{3 e ^{x} + 6} = 2

1 = 0. 5^{x}

2^{x - 1} = 8

9^{2 x - 1} = 2 7^{x + 2}

What is the answer to 10^x+10^{x+1}=22 ?
The answer to 10^x+10^{x+1}=22 is x=(ln(2))/(ln(10))
Solve for x: 10^x+10^{x+1}=22
The solution is x=(ln(2))/(ln(10))