What is the answer to 3^x-6^{3x+1}=0 ?

The answer to 3^x-6^{3x+1}=0 is x=-(ln(6))/(ln(72))

Solve for x: 3^x-6^{3x+1}=0

The solution is x=-(ln(6))/(ln(72))

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3^{x} - 6^{3 x + 1} = 0

x = - \frac{ln ( 6 )}{ln ( 72 )}

Decimal Notation

x = - 0.41896 \dots

Solution steps

3^{x} - 6^{3 x + 1} = 0

Add

6^{3 x + 1}

to both sides

3^{x} - 6^{3 x + 1} + 6^{3 x + 1} = 0 + 6^{3 x + 1}

Simplify

3^{x} = 6^{3 x + 1}

Apply exponent rules

x ln (3) = (3 x + 1) ln (6)

Solve

x ln (3) = (3 x + 1) ln (6) : x = - \frac{ln ( 6 )}{ln ( 72 )}

x = - \frac{ln ( 6 )}{ln ( 72 )}

1200000 = 600000 e^{0.05 t}

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

e^{2 x} = e^{7 x}

e^{4 x + 4} - e^{- x - 1} = 0

9^{- 3 x} \cdot 9^{x} = 27

What is the answer to 3^x-6^{3x+1}=0 ?
The answer to 3^x-6^{3x+1}=0 is x=-(ln(6))/(ln(72))
Solve for x: 3^x-6^{3x+1}=0
The solution is x=-(ln(6))/(ln(72))