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

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

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

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

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5^{2 x - 1} = 7^{x - 5}

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

Solution steps

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

Apply exponent rules

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

Solve

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

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

(\frac{1}{5})^{x} = 5

6^{x} - 4 \cdot 3^{x} = 0

\frac{3 e ^{x}}{( 3 + e ^{x} ) ^{2}} = 0

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

4 e^{x} = e^{4 x}

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