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Question

Find the value of $$x$$, if $$\log _{2}\left ( 4\times 3^{x}-6 \right )-\log _{2}\left ( 9^{x}-6 \right )=1$$


Solution

$$\log _{2}\left ( 4\times 3^{x}-6 \right )-\log _{2}\left ( 9^{x}-6 \right )=1$$.
$$\Rightarrow $$  $$\displaystyle \log _{2}\frac{4\times 3^{x}-6}{9^{x}-6}=1 \quad [\because \log a-\log b=\dfrac{a}{b}]$$
$$\Rightarrow $$   $$\displaystyle \frac{4\times 3^{x}-6}{9^{x}-6}=2$$
Substituting, $$3^{x}=y$$, we get
$$\Rightarrow $$   $$4y-6=2y^{2}-12$$  
$$\Rightarrow $$   $$y^{2}-2y-3=0$$
$$\Rightarrow $$   $$y=-1, 3$$
$$\Rightarrow $$   $$3^{x}=3$$
$$\Rightarrow $$   $$x=1$$
Ans: 1

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