According to the basic rules of $\log$, I'm solving both $\log$ terms as for first: base is $3$, $N$ is $9$ so exponent is calculated as $2$, and same for other term. But I'm confused with this '$x$'. Adding both terms $\log$ according to my logic will result in $4$. But I know I'm doing something wrong here, how to treat this $x$? Is B the correct answer?



Recall that

$$\log_3 (9\cdot x)=\log_3 9+\log_3 x$$

then what about $\log_2 (4\cdot x)$?

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  • 1
    $\begingroup$ @MichaelBurr Looking at the given choices I think that this hint can be useful. $\endgroup$ – user Nov 30 '18 at 14:36
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    $\begingroup$ Got it!!! The answer is B. Thank you so much!! $\endgroup$ – shawn k Nov 30 '18 at 14:36
  • $\begingroup$ @shawnk Exactly! Well done! Bye $\endgroup$ – user Nov 30 '18 at 14:37

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