What is the answer to 4^{(7x-2)}=17^{(x+1)} ?

The answer to 4^{(7x-2)}=17^{(x+1)} is x=(ln(272))/(14ln(2)-ln(17))

Solve for x: 4^{(7x-2)}=17^{(x+1)}

The solution is x=(ln(272))/(14ln(2)-ln(17))

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4^{(7 x - 2)} = 1 7^{(x + 1)}

x = \frac{ln ( 272 )}{14 ln ( 2 ) - ln ( 17 )}

Decimal Notation

x = 0.81588 \dots

Solution steps

4^{(7 x - 2)} = 1 7^{(x + 1)}

Apply exponent rules

(7 x - 2) ln (4) = (x + 1) ln (17)

Solve

(7 x - 2) ln (4) = (x + 1) ln (17) : x = \frac{ln ( 272 )}{14 ln ( 2 ) - ln ( 17 )}

x = \frac{ln ( 272 )}{14 ln ( 2 ) - ln ( 17 )}

e^{x - 1} - x - 1 = 0

e^{3 x} - 3 e^{2 x} + 4 e^{x} - 4 = 0

3^{2 x} = 3^{x + 5}

e^{- 0.2 t} = 121

50 = 73 + (36 - 73) e^{- k (45)}

What is the answer to 4^{(7x-2)}=17^{(x+1)} ?
The answer to 4^{(7x-2)}=17^{(x+1)} is x=(ln(272))/(14ln(2)-ln(17))
Solve for x: 4^{(7x-2)}=17^{(x+1)}
The solution is x=(ln(272))/(14ln(2)-ln(17))