What is the answer to 7^{4x}=31^{3x+4} ?

The answer to 7^{4x}=31^{3x+4} is x=(4ln(31))/(4ln(7)-3ln(31))

Solve for x: 7^{4x}=31^{3x+4}

The solution is x=(4ln(31))/(4ln(7)-3ln(31))

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

x = \frac{4 ln ( 31 )}{4 ln ( 7 ) - 3 ln ( 31 )}

Solution steps

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

Apply exponent rules

4 x ln (7) = (3 x + 4) ln (31)

Solve

4 x ln (7) = (3 x + 4) ln (31) : x = \frac{4 ln ( 31 )}{4 ln ( 7 ) - 3 ln ( 31 )}

x = \frac{4 ln ( 31 )}{4 ln ( 7 ) - 3 ln ( 31 )}

2^{x} = 162

(\frac{1}{4})^{2 x} = 3 2^{4 x - 2}

3^{2 x} = 243

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

5 (2)^{x + 1} = 9

What is the answer to 7^{4x}=31^{3x+4} ?
The answer to 7^{4x}=31^{3x+4} is x=(4ln(31))/(4ln(7)-3ln(31))
Solve for x: 7^{4x}=31^{3x+4}
The solution is x=(4ln(31))/(4ln(7)-3ln(31))