Find $x$ , if $9 \times 3^{x} = (27)^{2 x - 3}$

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Solution

Given equation: $9 \times 3^{x} = (27)^{2 x - 3} \Rightarrow 3^{2} \times 3^{x} = (3 \times 3 \times 3)^{2 x - 3} \Rightarrow 3^{x + 2} = (3)^{3 (2 x - 3)} [∵ x^{a} \times x^{b} = x^{a + b}, (x^{a})^{b} = x^{a \times b}] \Rightarrow 3^{x + 2} = (3)^{6 x - 9}$
Since, the bases are same, compare their exponents,
$\Rightarrow x + 2 = 6 x - 9$
$\Rightarrow 6 x - x = 9 + 2$
$\Rightarrow 5 x = 11$
$\Rightarrow x = \frac{11}{5}$
$\Rightarrow x = 2 \frac{1}{5}$
Hence, the value of $x = 2 \frac{1}{5}$ .

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