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 )}

3^{2 x - 1} = 93

5^{x} = 5^{2}

\frac{1}{2 ^{x + 4}} = 8

2 = 1.0 2^{x}

2 5^{x} \cdot 5^{x^{2}} = 62 5^{2}

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))