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

The answer to 5(7^{2x-1})=7(5^{x+2}) is x=(ln(245))/(2ln(7)-ln(5))

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

The solution is x=(ln(245))/(2ln(7)-ln(5))

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

x = \frac{ln ( 245 )}{2 ln ( 7 ) - ln ( 5 )}

Decimal Notation

x = 2.41031 \dots

Solution steps

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

Apply exponent rules

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

Solve

ln (5) + (2 x - 1) ln (7) = ln (7) + (x + 2) ln (5) : x = \frac{ln ( 245 )}{2 ln ( 7 ) - ln ( 5 )}

x = \frac{ln ( 245 )}{2 ln ( 7 ) - ln ( 5 )}

π^{1 - x} = e^{x}

4^{x} + 4^{- x} = \frac{5}{2}

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

2^{x} = \frac{1}{16}

10 0^{x - 1} = 100 0^{2 x}

What is the answer to 5(7^{2x-1})=7(5^{x+2}) ?
The answer to 5(7^{2x-1})=7(5^{x+2}) is x=(ln(245))/(2ln(7)-ln(5))
Solve for x: 5(7^{2x-1})=7(5^{x+2})
The solution is x=(ln(245))/(2ln(7)-ln(5))