What is the answer to 6(10)^x+(10)^{x-1}=2 ?

The answer to 6(10)^x+(10)^{x-1}=2 is x=(ln(2/61))/(ln(10))+1

Solve for x: 6(10)^x+(10)^{x-1}=2

The solution is x=(ln(2/61))/(ln(10))+1

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6 (10)^{x} + (10)^{x - 1} = 2

x = \frac{ln ( \frac{2}{61} )}{ln ( 10 )} + 1

Decimal Notation

x = - 0.48429 \dots

Solution steps

6 (10)^{x} + (10)^{x - 1} = 2

Factor

6 \cdot 1 0^{x} + 1 0^{x - 1} : 61 \cdot 1 0^{x - 1}

61 \cdot 1 0^{x - 1} = 2

Divide both sides by

61

1 0^{x - 1} = \frac{2}{61}

Apply exponent rules

(x - 1) ln (10) = ln (\frac{2}{61})

Solve

(x - 1) ln (10) = ln (\frac{2}{61}) : x = \frac{ln ( \frac{2}{61} )}{ln ( 10 )} + 1

x = \frac{ln ( \frac{2}{61} )}{ln ( 10 )} + 1

x, 5^{2 x - 5} = 125

e^{0.144 x} = 14

e^{2 x} - 14 = 5 e^{x}

24 = 1 0^{x}

2^{2 x - 1} = 4^{3 x + 2}

What is the answer to 6(10)^x+(10)^{x-1}=2 ?
The answer to 6(10)^x+(10)^{x-1}=2 is x=(ln(2/61))/(ln(10))+1
Solve for x: 6(10)^x+(10)^{x-1}=2
The solution is x=(ln(2/61))/(ln(10))+1