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

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

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

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

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

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

Decimal

x = 3.81884 \dots

Solution steps

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

Apply exponent rules

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

Solve

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

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

3 6^{x - 5} = \frac{1}{6}

25 = \frac{100 ( 0.5 ) ^{t}}{5730}

2 (7^{x}) = 686

9^{- x} = 27

81 (3^{x^{2} + 5 x}) = 243

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