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

The answer to 7^{3x-1}=5^{x+1} is x=(ln(35))/(3ln(7)-ln(5))

Solve for x: 7^{3x-1}=5^{x+1}

The solution is x=(ln(35))/(3ln(7)-ln(5))

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7^{3 x - 1} = 5^{x + 1}

x = \frac{ln ( 35 )}{3 ln ( 7 ) - ln ( 5 )}

Decimal Notation

x = 0.84084 \dots

Solution steps

7^{3 x - 1} = 5^{x + 1}

Apply exponent rules

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

Solve

(3 x - 1) ln (7) = (x + 1) ln (5) : x = \frac{ln ( 35 )}{3 ln ( 7 ) - ln ( 5 )}

x = \frac{ln ( 35 )}{3 ln ( 7 ) - ln ( 5 )}

2^{2 x - 4} = 8^{4}

1 6^{2 x - 2} = 2^{3 x + 3}

(2 \times (- 5))^{x^{2} - 5 x + 4} = 1

x^{x} = 6

3^{3 x + 5} = 3 \cdot 9^{x + 3}

What is the answer to 7^{3x-1}=5^{x+1} ?
The answer to 7^{3x-1}=5^{x+1} is x=(ln(35))/(3ln(7)-ln(5))
Solve for x: 7^{3x-1}=5^{x+1}
The solution is x=(ln(35))/(3ln(7)-ln(5))