What is (sqrt(3))^{1-x}\times 9^{1-2x} ?

The solution to (sqrt(3))^{1-x}\times 9^{1-2x} is 3^{(-9x+1)/2+2}

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simplify (sqrt(3))^{1-x}\times 9^{1-2x}

simplify

(3)^{1 - x} \cdot 9^{1 - 2 x}

3^{\frac{- 9 x + 1}{2} + 2}

Solution steps

(3)^{1 - x} \cdot 9^{1 - 2 x}

(3)^{1 - x} : 3^{\frac{1 - x}{2}}

= 3^{\frac{1 - x}{2}} \cdot 9^{1 - 2 x}

Factor integer

9 = 3^{2}

= 3^{\frac{1 - x}{2}} (3^{2})^{1 - 2 x}

Apply exponent rule:

(a^{b})^{c} = a^{b c}

(3^{2})^{1 - 2 x} = 3^{2 (1 - 2 x)}

= 3^{\frac{1 - x}{2}} \cdot 3^{2 (1 - 2 x)}

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

3^{\frac{1 - x}{2}} \cdot 3^{2 (1 - 2 x)} = 3^{\frac{- x + 1}{2} + 2 (- 2 x + 1)}

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

Expand

\frac{1 - x}{2} + 2 (1 - 2 x) : \frac{1 - 9 x}{2} + 2

= 3^{\frac{- 9 x + 1}{2} + 2}

e^{- 4}

e^{- 3.75}

1. 2^{- 1}

3500 \times e^{- 0.0001 x}

What is (sqrt(3))^{1-x}\times 9^{1-2x} ?
The solution to (sqrt(3))^{1-x}\times 9^{1-2x} is 3^{(-9x+1)/2+2}