Quantcast

Got Homework?

Connect with other students for help. It's a free community.

  • across
    MIT Grad Student
    Online now
  • laura*
    Helped 1,000 students
    Online now
  • Hero
    College Math Guru
    Online now

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

dmartinez130

Solve log4 32=x+2. Can someone please help me solve this?

  • one year ago
  • one year ago

  • This Question is Closed
  1. dmartinez130
    Best Response
    You've already chosen the best response.
    Medals 0

    4 is the base of log.

    • one year ago
  2. dmartinez130
    Best Response
    You've already chosen the best response.
    Medals 0

    \[\log _{4}32=x+2\]

    • one year ago
  3. myininaya
    Best Response
    You've already chosen the best response.
    Medals 0

    To solve for x Subtract 2 on both sides

    • one year ago
  4. dmartinez130
    Best Response
    You've already chosen the best response.
    Medals 0

    So, log4 30=x?

    • one year ago
  5. Ravus
    Best Response
    You've already chosen the best response.
    Medals 1

    Change it to the exponential form. Tell me what you see

    • one year ago
  6. Ravus
    Best Response
    You've already chosen the best response.
    Medals 1

    I think my way is easier :p try this first ^

    • one year ago
  7. Ravus
    Best Response
    You've already chosen the best response.
    Medals 1

    4^(x+2)=32 do you know how to get that?

    • one year ago
  8. dmartinez130
    Best Response
    You've already chosen the best response.
    Medals 0

    No I don't :/

    • one year ago
  9. myininaya
    Best Response
    You've already chosen the best response.
    Medals 0

    \[\log_432=x+2\] -2 -2 \[\log_432-2=x\]

    • one year ago
  10. Ravus
    Best Response
    You've already chosen the best response.
    Medals 1

    The whole point of logs, is that they were "made" to "Solve for" an exponent. So if I had an expression that was 2^x=100, the only way to "find x" is to use logarithms. How? You can re-write any exponential equation to look like a logarithm, and vise versa.

    • one year ago
  11. Ravus
    Best Response
    You've already chosen the best response.
    Medals 1

    To go from log to exponential, you need to write the "answer" as the EXPONENT of the BASE. 4 is the base of the log, so write x+2 as the exponent of 4. 4^(x+2)

    • one year ago
  12. Ravus
    Best Response
    You've already chosen the best response.
    Medals 1

    and what does that equal? Well you only have one number left; the number you were taking the log of. 32. This is a basic skill you need to learn :) Writing logs as exponentials, and writiing exponentials as logs.

    • one year ago
  13. myininaya
    Best Response
    You've already chosen the best response.
    Medals 0

    Now if you want to simplify \[\log_432\] I would say let g equal that so we have \[g=\log_432\] writing in exponential form we get \[4^g=32\] g=3 doesn't work b/c 4(4)(4)=16(4)=64 g=3/2 doesn't work b/c (4)^(3/2)=2^3=8 g=5/2 works b/c (4)^(5/2)=2^5=32 Booyah! :) So that means \[\log_4(32)=\frac{5}{2}\] So what does that mean x equals or simplifies to?

    • one year ago
  14. myininaya
    Best Response
    You've already chosen the best response.
    Medals 0

    You could just use your calculator though instead of that method right there sometimes that method will be too difficult to use

    • one year ago
  15. myininaya
    Best Response
    You've already chosen the best response.
    Medals 0

    A good formula to remember is the change of base formula \[\frac{\ln(a)}{\ln(b)}=\log_b(a)\]

    • one year ago
  16. Ravus
    Best Response
    You've already chosen the best response.
    Medals 1

    \[4^{x+2}=32 \] So how do you solve this? There's a trick you'll probably have to often use: you need express both numbers as the same base! You can probably tell that 4 is simply 2^2, and 32 is 2^5 That's what I mean by "the same base" So once you do that, you've got this: \[2^{2(x+2)}=2^{5}\] Once you have something looking like that, you can solve for x! You simply "ignore" the "big" 2s, and you get 2(x+2)=5 solve for x now, and you get 0.5, or 1/2 :)

    • one year ago
  17. myininaya
    Best Response
    You've already chosen the best response.
    Medals 0

    using my way we get \[\frac{5}{2}-2=x\]

    • one year ago
  18. myininaya
    Best Response
    You've already chosen the best response.
    Medals 0

    Remember since \[\log_4(32)=\frac{5}{2}\]

    • one year ago
  19. Ravus
    Best Response
    You've already chosen the best response.
    Medals 1

    The trick to these is: 1. writing it in exponential form (learn to do this!! you'll use it all the time) 2. Expressing both sides of your new (but equilivant to the log expression) equation with the same bases. 3."zooming in" on the exponents, ignoring the "big numbers" (which are now the same) and solving for x like a regular algebra problem.

    • one year ago
  20. Ravus
    Best Response
    You've already chosen the best response.
    Medals 1

    myininaya i like your solution, it's just that I'm close to the asker's age, I think, and I remember that this is the particular way they want these questions done, even if it is a drag , lol

    • one year ago
  21. Ravus
    Best Response
    You've already chosen the best response.
    Medals 1

    they=the teachers

    • one year ago
    • Attachments:

See more questions >>>

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.