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 )}

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 )}

2^{2 x + 3} = 8

x, e^{r x} = 0

e^{9 - 4 x} = 8

e^{3 x} = ln (4)

\frac{1 0 ^{12}}{1 0 ^{y}} = 1 0^{4}

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))