What is the answer to 4^{x+1}=3^x ?

The answer to 4^{x+1}=3^x is x=-(2ln(2))/(2ln(2)-ln(3))

Solve for x: 4^{x+1}=3^x

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

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4^{x + 1} = 3^{x}

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

Solution steps

4^{x + 1} = 3^{x}

Apply exponent rules

(x + 1) ln (4) = x ln (3)

Solve

(x + 1) ln (4) = x ln (3) : x = - \frac{2 ln ( 2 )}{2 ln ( 2 ) - ln ( 3 )}

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

8^{3 x + 1} = 3 2^{x}

9^{x} = 729

e^{z} + 1 = 0

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

9^{x - 2} = 27

What is the answer to 4^{x+1}=3^x ?
The answer to 4^{x+1}=3^x is x=-(2ln(2))/(2ln(2)-ln(3))
Solve for x: 4^{x+1}=3^x
The solution is x=-(2ln(2))/(2ln(2)-ln(3))