What is the answer to 3(2^x)=4^{x+1} ?

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

Solve for x: 3(2^x)=4^{x+1}

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

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3 (2^{x}) = 4^{x + 1}

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

Decimal Notation

x = - 0.41503 \dots

Solution steps

3 (2^{x}) = 4^{x + 1}

Apply exponent rules

ln (3) + x ln (2) = 2 (x + 1) ln (2)

Solve

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

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

9^{2} = 3^{2 x + 1}

4^{x} = 18

e^{5 x} = 9

\frac{1}{27} = 3^{4 m - 1}

3^{x} = 5^{2 x - 1}

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