What is the answer to (3^x)(4^{2x+1})=6^{x+2} ?

The answer to (3^x)(4^{2x+1})=6^{x+2} is x=(2ln(3))/(3ln(2))

Solve for x: (3^x)(4^{2x+1})=6^{x+2}

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

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(3^{x}) (4^{2 x + 1}) = 6^{x + 2}

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

Decimal Notation

x = 1.05664 \dots

Solution steps

(3^{x}) (4^{2 x + 1}) = 6^{x + 2}

Apply exponent rules

(x + lo g_{3} (4) (2 x + 1)) ln (3) = (x + 2) ln (6)

Solve

(x + lo g_{3} (4) (2 x + 1)) ln (3) = (x + 2) ln (6) : x = \frac{2 ln ( 3 )}{3 ln ( 2 )}

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

\frac{1 1 ^{x}}{1 1 ^{3}} = 1 1^{2}

5^{2 x} - 5^{x} - 12 = 0

e^{x} + e^{x} x = 0

2^{x} = \frac{2 ^{a} 2 ^{b}}{2 ^{c}}

1 0^{x + 1} = e^{2 x}

What is the answer to (3^x)(4^{2x+1})=6^{x+2} ?
The answer to (3^x)(4^{2x+1})=6^{x+2} is x=(2ln(3))/(3ln(2))
Solve for x: (3^x)(4^{2x+1})=6^{x+2}
The solution is x=(2ln(3))/(3ln(2))