What is the answer to 10^x+2*10^{x+1}=42 ?

The answer to 10^x+2*10^{x+1}=42 is x=(ln(2))/(ln(10))

Solve for x: 10^x+2*10^{x+1}=42

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

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1 0^{x} + 2 \cdot 1 0^{x + 1} = 42

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

Decimal Notation

x = 0.30102 \dots

Solution steps

1 0^{x} + 2 \cdot 1 0^{x + 1} = 42

Factor

1 0^{x} + 2 \cdot 1 0^{x + 1} : 21 \cdot 1 0^{x}

21 \cdot 1 0^{x} = 42

Divide both sides by

21

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 )}

\frac{3}{4} = (\frac{4}{3})^{- x}

y, x = b^{y}

38. 5^{x} = 6. 5^{(x - 2)}

(- 2)^{3^{2}} - (- 2)^{2^{x}} = 0

3^{2 (x - 1)} = 2^{x - 3}

What is the answer to 10^x+2*10^{x+1}=42 ?
The answer to 10^x+2*10^{x+1}=42 is x=(ln(2))/(ln(10))
Solve for x: 10^x+2*10^{x+1}=42
The solution is x=(ln(2))/(ln(10))