Quantcast

A community for students. Sign up today!

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

cayssaday

  • 9 months ago

Can someone pleassssse help me with this problem? :) Solve for x. Enter only a number. 3 = log8 + 3logx HINT: If loga = 3, then a = 103.

  • This Question is Closed
  1. MrNood
    • 9 months ago
    Best Response
    You've already chosen the best response.
    Medals 0

    log8 is just a number so 3-log8 = 3logx or logx = (3-log8)/3 try it from there...

  2. AkashdeepDeb
    • 9 months ago
    Best Response
    You've already chosen the best response.
    Medals 0

    a = 10^3 and not 103. 3 - log 8 = 3 * log x We know that 3 = log 10^3 So log 10^3 - log 8 = log 10^3 * log x Log rules tell us that log p - log q = log (p/q) Thus the resulting equation is log (10^3 / 8) = log 10^3 * log x Can you figure out the solution from here? :)

  3. cayssaday
    • 9 months ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I thinkkk so, thank you so much !

  4. surjithayer
    • 9 months ago
    Best Response
    You've already chosen the best response.
    Medals 1

    \[3=\log 8++\log x^3=\log (8x^3),8x^3=10^3,(2x)^3=10^3\]

  5. surjithayer
    • 9 months ago
    Best Response
    You've already chosen the best response.
    Medals 1

    2x=10 x=?

  6. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Ask a Question
Find 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
  • 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.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.