What is the answer to 7^{2x}=2(5^{3x+1}) ?

The answer to 7^{2x}=2(5^{3x+1}) is x=(ln(10))/(2ln(7)-3ln(5))

Solve for x: 7^{2x}=2(5^{3x+1})

The solution is x=(ln(10))/(2ln(7)-3ln(5))

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7^{2 x} = 2 (5^{3 x + 1})

x = \frac{ln ( 10 )}{2 ln ( 7 ) - 3 ln ( 5 )}

Solution steps

7^{2 x} = 2 (5^{3 x + 1})

Apply exponent rules

2 x ln (7) = ln (2) + (3 x + 1) ln (5)

Solve

2 x ln (7) = ln (2) + (3 x + 1) ln (5) : x = \frac{ln ( 10 )}{2 ln ( 7 ) - 3 ln ( 5 )}

x = \frac{ln ( 10 )}{2 ln ( 7 ) - 3 ln ( 5 )}

e^{3 x} = 0

4^{l o g_{4} (x) + l o g_{4} (2)} = 3

5^{x} = 30

9^{x} = (\frac{1}{27})^{x - 2}

9^{x + 1} = 2 7^{3 x - 4}

What is the answer to 7^{2x}=2(5^{3x+1}) ?
The answer to 7^{2x}=2(5^{3x+1}) is x=(ln(10))/(2ln(7)-3ln(5))
Solve for x: 7^{2x}=2(5^{3x+1})
The solution is x=(ln(10))/(2ln(7)-3ln(5))