What is the answer to 1050^x+1050^{x+1}+1050^{x+2}=2021 ?

The answer to 1050^x+1050^{x+1}+1050^{x+2}=2021 is x=(ln(2021/1103551))/(ln(1050))

Solve for x: 1050^x+1050^{x+1}+1050^{x+2}=2021

The solution is x=(ln(2021/1103551))/(ln(1050))

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105 0^{x} + 105 0^{x + 1} + 105 0^{x + 2} = 2021

x = \frac{ln ( \frac{2021}{1103551} )}{ln ( 1050 )}

Decimal Notation

x = - 0.90600 \dots

Solution steps

105 0^{x} + 105 0^{x + 1} + 105 0^{x + 2} = 2021

Factor

105 0^{x} + 105 0^{x + 1} + 105 0^{x + 2} : 1103551 \cdot 105 0^{x}

1103551 \cdot 105 0^{x} = 2021

Divide both sides by

1103551

105 0^{x} = \frac{2021}{1103551}

Apply exponent rules

x ln (1050) = ln (\frac{2021}{1103551})

Solve

x ln (1050) = ln (\frac{2021}{1103551}) : x = \frac{ln ( \frac{2021}{1103551} )}{ln ( 1050 )}

x = \frac{ln ( \frac{2021}{1103551} )}{ln ( 1050 )}

4^{x} = 26

4^{x} = 21

4 \times 7^{x} = 4

- 3^{x} = - 3^{13}

9^{x} - 43 \times 3^{x - 1} + 1 = 0

What is the answer to 1050^x+1050^{x+1}+1050^{x+2}=2021 ?
The answer to 1050^x+1050^{x+1}+1050^{x+2}=2021 is x=(ln(2021/1103551))/(ln(1050))
Solve for x: 1050^x+1050^{x+1}+1050^{x+2}=2021
The solution is x=(ln(2021/1103551))/(ln(1050))