A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

marcelie

  • one year ago

Help please !!! Solve the following with a calculator. 2log (100^-3)

  • This Question is Closed
  1. marcelie
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    |dw:1444009782868:dw|

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

    -12

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

    how you get that

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

    Just put it into your calculator or google :)

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

    lol z.. i wanna know know how solve it by hand.

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

    apply the power rule

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

    power rule \[\large\rm log_b x^y = y \log_b x\]

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

    \[\huge\rm 2 \log (100^{-3})\] number at front of log becomes exponent of (100^{-3})\[\large\rm log (100^{-3})^2\] now multiply the exponents \[\rm (X^m)^n=x^{m \times n}\] and then negative exponent rule \[\large\rm x^{-m}=\frac{ 1 }{ x^m }\] then there are two ways to find final answer 1st) use change of base formula and then calculator or 2nd) set it equal to x to solve by hand

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

    make sense ? take ur time to read that :=)

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

    one second let me attach the picture

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

    1 Attachment
  12. Nnesha
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    \(\huge\color{green}{\checkmark}\)

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

    gimme a sec i'll check it again

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

    looks right

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

    aiight got it :) do you know any website that has log formulas

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

    quotient rule\[\large\rm log_b x - \log_b y = \log_b \frac{ x }{ y}\] to condense you can change subtraction to division product rule \[\large\rm log_b x + \log_b y = \log_b( x \times y )\] addition ----> multiplication power rule \[\large\rm log_b x^y = y \log_b x\]

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

    heheh change of base formula \[\huge\rm \log_b a = \frac{ \log a }{ \log b }\]

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

    http://www.rapidtables.com/math/algebra/Logarithm.htm

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

    \[\begin{align} 2\log(100^{-3}) &=2\log\left(\frac{1}{100^3}\right) \\& = 2[\log(1) -\log(100^3)] \\ &=2[0-\log(10^6)] \\&= 2[-6\log(10)] \\&= 2(-6(1))\\&= -12 \end{align} \]

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

    That's another approach you could have used as well :P

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

    ah i see it :) Thank you ladies :)

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

    @Nnesha what do you think?

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

    nice!

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

    or after this if you don't want to use change of base formula set it equal to x \[\huge\rm log (100^{-6}) =x\] now change log to an exponential form log= log base 10 \[10^x =100^{-6}\] rewrite 100^{-16} in terms of base 10 \[10^x =10^{2 \times -6}\] bases are same you can cancel \[\cancel{10}^x =\cancel{10}^{2 \times -6}\] left with x= 2 times -6 = -12 :=)

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

    |dw:1444012127376:dw| that's all you need to know for logs or u can visit that website :=)

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

    \[\large \log_{10}(100^{-6}) = x\]\[10^{\large \log_{10}(100^{-6})} = 10^x\]\[\large 100^{-6} = 10^{x} \qquad \qquad \qquad 10^{-12} \equiv 100^{-6} \]\[\large 10^{-12} = 10^{x}\]\[\therefore -12 = x\]

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

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

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.