If $(3^{x^{2}}) (9^{x}) (3) = 27$ and $x > 0$ , compute the value of $x$ .

0.268

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0.414

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0.732

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1.414

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Solution

The correct option is D $0.732$
Given, $(3^{x^{2}}) (9^{x}) (3) = 27$
Taking LHS:
$(3^{x^{2}}) (9^{x}) (3)$
$\Rightarrow (3^{x^{2}}) (3^{2 x}) (3^{1}) \Rightarrow 3^{x^{2} + 2 x + 1}$
Now taking RHS: = $27 = 3^{3}$
Equating both, we get
$3^{x^{2} + 2 x + 1} = 3^{3}$
Both bases are equal, so powers must be equal.
$\Rightarrow x^{2} + 2 x + 1 = 3 \Rightarrow x^{2} + 2 x - 2 = 0 \Rightarrow x = 0.7320508075, - 2.73205080$

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Similar questions

0.414 - 0.732

If the radius ratio is between 0.414 - 0.732, what will be the type of void present? What will be the coordination number?

\sqrt{2} = 1.414

\sqrt{\frac{(\sqrt{2} - 1)}{(\sqrt{2} + 1)}}

\sqrt{2} = 1.414 then \sqrt{\frac{(\sqrt{2} - 1)}{(\sqrt{2} + 1)}} = ?

If $\sqrt{2} = 1.414 t h e n \sqrt{\frac{(\sqrt{2} - 1)}{(\sqrt{2} + 1)}} = ?$

(a) 0.207 (b) 2.414 (c) 0.414 (d) 0.621

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