What is the answer to 20(7^{2x})=40(5^{3x+1}) ?

The answer to 20(7^{2x})=40(5^{3x+1}) is x=(ln(10))/(2ln(7)-3ln(5))

Solve for x: 20(7^{2x})=40(5^{3x+1})

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

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20 (7^{2 x}) = 40 (5^{3 x + 1})

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

Decimal Notation

x = - 2.45873 \dots

Solution steps

20 (7^{2 x}) = 40 (5^{3 x + 1})

Apply exponent rules

ln (20) + 2 x ln (7) = ln (40) + (3 x + 1) ln (5)

Solve

ln (20) + 2 x ln (7) = ln (40) + (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 )}

7^{x + 2} = 98

x, e^{a x} = C e^{b x}

5^{x} - 5^{5 - x} = 150

e^{5 - 3 x} = 10

y, x = e^{y}

What is the answer to 20(7^{2x})=40(5^{3x+1}) ?
The answer to 20(7^{2x})=40(5^{3x+1}) is x=(ln(10))/(2ln(7)-3ln(5))
Solve for x: 20(7^{2x})=40(5^{3x+1})
The solution is x=(ln(10))/(2ln(7)-3ln(5))