A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 4 years ago
Log question
anonymous
 4 years ago
Log question

This Question is Closed

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[25^{\log_{4}x }5^{(\log _{16}x ^{2})+1}=\log _{\sqrt{3}}9\sqrt{3}25^{\log _{16}x}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\Large{25^{\log_{4}x }5^{(\log _{16}x ^{2})+1}=\log _{\sqrt{3}}9\sqrt{3}25^{\log _{16}x}}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0doe that help solve the problem

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0No I just don't see how you can be doing these sort of things in grade 10

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I wouldn't know where to start for this, but I can guess your answer will be a multiple of 4. If not, it could get ugly. :P

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Okay, I think x is 4, which makes the overall equation 0 = 0.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\large{25^{\log_{4}4 }5^{(\log _{16}4 ^{2})+1}=\log _{\sqrt{3}}9\sqrt{3}25^{\log _{16}4}}\] \[\large{255^{2}=\log _{\sqrt{3}}\sqrt{243}25^{0.5}}\] \[2525 = 0 =\log({\frac{\sqrt{243}}{\sqrt{3}})5} = 5  5 = 0\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0How did you realise it had to be a multiple of 4?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Because all the bases of the logs were multiples of 4 (either 4 or 16), so I knew anything that wasn't a multiple of 4 would be too messy.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0And when \(x = 4\) gave me \(\log_{4}4\) and \(\log_{16}16\) on the left side, I figured I was on to something,

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Makes a lot of sense.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I did notice one mistake in my explanation, however. It should say: \[\frac{\log{\sqrt{243}}}{\log{\sqrt{3}}}\] not \[\log({\frac{\sqrt{243}}{\sqrt{3}}})\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Okay, I've tried to include as many parts of my process as I could. If you have any questions about what I did, @ArkGoLucky, let me know. \[\Large{25^{\log_{4}x }5^{(\log _{16}x ^{2})+1}=\log _{\sqrt{3}}9\sqrt{3}25^{\log _{16}x}}\] \[\Large{25^{\log_{4}x }25^{\log _{16}x ^{2}}=525^{\log _{16}x}}\] \[\Large{25^{2 \log_{16}x }25^{\log _{16}x ^{2}}=525^{\log _{16}x}}\] \[\Large{25^{2 \log_{16}x }25^{\log _{16}x ^{2}}+25^{\log _{16}x}=5}\] \[\Large{25^{2 \log_{16}x }25^{2\log _{16}x}+25^{\log _{16}x}=5}\] \[\Large{25^{\log _{16}x}=5}\] \[\Large{5^{2 \log _{16}x}=5}\] \[\Large{2 \log _{16}x=1}\] \[\Large{\log _{16}x=0.5}\] \[\Large{x = 16^{0.5}=\sqrt{16} = 4}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0How did you get to the second step lol

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\Large{\log_{\sqrt{3}}9 \sqrt{3} = 5}\] \[\Large{5^{(\log_{16}x^{2})+1} = 5^{\log_{16}x^{2}} \times 5^{1} = 25^{\log_{16}x^{2}}}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Ah I didn't think it would come out to be a whole number!! Great job on it.
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.