Solve the following equation:
${log}_{2} ({4.3}^{x} - 6) - {log}_{2} (9^{x} - 6) = 1$

x = 4

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x = 3

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x = 2

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x = 1

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Solution

The correct option is D $x = 1$
${l o g}_{2} ({4.3}^{x} - 6) - {l o g}_{2} (9^{x} - 6) = 1 \Rightarrow {l o g}_{2} (\frac{4. 3^{x} - 6}{9^{x} - 6}) = 1 \Rightarrow \frac{{4.3}^{x} - 6}{{(3^{x})}^{2} - 6} = 2 \Rightarrow {(3^{x})}^{2} - 6 = {2.3}^{x} - 3 \Rightarrow {{(3}^{x})}^{2} - {2.3}^{x} - 3 = 0$

Therefore, $3^{x} = - 1$ or $3^{x} = 3$
But, $3^{x} \neq - 1$

Therefore, $x = 1.$

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Solve the following equation:log2(4.3x−6)−log2(9x−6)=1

The correct option is D x=1log2(4.3x−6)−log2(9x−6)=1⇒log2(4.3x−69x−6)=1⇒4.3x−6(3x)2−6=2⇒(3x)2−6=2.3x−3⇒(3x)2−2.3x−3=0Therefore, 3x=−1 or 3x=3But, 3x≠−1Therefore, x=1.

Solve the following equation:
${log}_{2} ({4.3}^{x} - 6) - {log}_{2} (9^{x} - 6) = 1$