What is the answer to 3^x=7^{4x+1} ?

The answer to 3^x=7^{4x+1} is x=(ln(7))/(ln(3)-4ln(7))

Solve for x: 3^x=7^{4x+1}

The solution is x=(ln(7))/(ln(3)-4ln(7))

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3^{x} = 7^{4 x + 1}

x = \frac{ln ( 7 )}{ln ( 3 ) - 4 ln ( 7 )}

Decimal Notation

x = - 0.29108 \dots

Solution steps

3^{x} = 7^{4 x + 1}

Apply exponent rules

x ln (3) = (4 x + 1) ln (7)

Solve

x ln (3) = (4 x + 1) ln (7) : x = \frac{ln ( 7 )}{ln ( 3 ) - 4 ln ( 7 )}

x = \frac{ln ( 7 )}{ln ( 3 ) - 4 ln ( 7 )}

\frac{e ^{x} + e ^{- x}}{2} = 3

8^{x} = \frac{2 ^{56} - 4 ^{26}}{30}

98 e^{\frac{x}{2}} = 490

1 0^{x - 2} = 1000

3^{5 x + 1} = 27

What is the answer to 3^x=7^{4x+1} ?
The answer to 3^x=7^{4x+1} is x=(ln(7))/(ln(3)-4ln(7))
Solve for x: 3^x=7^{4x+1}
The solution is x=(ln(7))/(ln(3)-4ln(7))