1
$\begingroup$

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?

$\endgroup$
3
$\begingroup$

HINT

Recall that

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

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

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ @MichaelBurr Looking at the given choices I think that this hint can be useful. $\endgroup$ – user Nov 30 '18 at 14:36
  • 2
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.