What is the answer to 4*2^{2t}-10*6^t*3^{2t}=0 ?

The answer to 4*2^{2t}-10*6^t*3^{2t}=0 is t=(ln(2/5))/(3ln(3)-ln(2))

Solve for t: 4*2^{2t}-10*6^t*3^{2t}=0

The solution is t=(ln(2/5))/(3ln(3)-ln(2))

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4 \cdot 2^{2 t} - 10 \cdot 6^{t} \cdot 3^{2 t} = 0

t = \frac{ln ( \frac{2}{5} )}{3 ln ( 3 ) - ln ( 2 )}

Decimal Notation

t = - 0.35205 \dots

Solution steps

4 \cdot 2^{2 t} - 10 \cdot 6^{t} \cdot 3^{2 t} = 0

Add

10 \cdot 6^{t} \cdot 3^{2 t}

to both sides

4 \cdot 2^{2 t} - 10 \cdot 6^{t} \cdot 3^{2 t} + 10 \cdot 6^{t} \cdot 3^{2 t} = 0 + 10 \cdot 6^{t} \cdot 3^{2 t}

Simplify

4 \cdot 2^{2 t} = 10 \cdot 6^{t} \cdot 3^{2 t}

Divide both sides by

2^{2 t}

\frac{4 \cdot 2 ^{2 t}}{2 ^{2 t}} = \frac{10 \cdot 6 ^{t} \cdot 3 ^{2 t}}{2 ^{2 t}}

Simplify

\frac{10 \cdot 6 ^{t} \cdot 3 ^{2 t}}{2 ^{2 t}} : 5 \cdot 2 7^{t} \cdot 2^{1 - t}

4 = 5 \cdot 2 7^{t} \cdot 2^{1 - t}

Apply exponent rules

(lo g_{2} (27) t + 1 - t) ln (2) = ln (\frac{4}{5})

Solve

(lo g_{2} (27) t + 1 - t) ln (2) = ln (\frac{4}{5}) : t = \frac{ln ( \frac{2}{5} )}{3 ln ( 3 ) - ln ( 2 )}

t = \frac{ln ( \frac{2}{5} )}{3 ln ( 3 ) - ln ( 2 )}

64 \cdot 1 6^{- 3 x} = 1 6^{- 3 x - 2}

4^{X + 1} - 3 \cdot 2^{X + 1} = 0

e^{2 x} - 2 e^{x} = 8

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

What is the answer to 4*2^{2t}-10*6^t*3^{2t}=0 ?
The answer to 4*2^{2t}-10*6^t*3^{2t}=0 is t=(ln(2/5))/(3ln(3)-ln(2))
Solve for t: 4*2^{2t}-10*6^t*3^{2t}=0
The solution is t=(ln(2/5))/(3ln(3)-ln(2))