Using the properties of logarithms: multiple steps (video) | Khan Academy (2024)

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  • pollackmarla

    12 years agoPosted 12 years ago. Direct link to pollackmarla's post “What do we do if there ar...”

    What do we do if there are three parts to subtract? All three logs have the same base.

    (11 votes)

    • financeBoy

      12 years agoPosted 12 years ago. Direct link to financeBoy's post “then you just divide divi...”

      Using the properties of logarithms: multiple steps (video) | Khan Academy (4)

      Using the properties of logarithms: multiple steps (video) | Khan Academy (5)

      then you just divide divide it 2 times...for example:
      log(3/2)/3... what you do: log3 - log 2 - log 3

      (25 votes)

  • Help Please

    12 years agoPosted 12 years ago. Direct link to Help Please's post “At 0:20 can we move the x...”

    At

    Using the properties of logarithms: multiple steps (video) | Khan Academy (7) 0:20

    can we move the x out in front first, instead of applying the quotient log property first? And then apply the quotient log property?

    If not...is there a certain order in which log properties should be applied (like BEDMAS)??? Thanks in advance :)

    (10 votes)

    • Sophie Reisinger

      12 years agoPosted 12 years ago. Direct link to Sophie Reisinger's post “You can't move the x in f...”

      Using the properties of logarithms: multiple steps (video) | Khan Academy (10)

      You can't move the x in front first, because not the whole factor is to the xth power. if we had "log(25/y)^x" we could. otherwise the whole term, so both log25 and log5 would be times 5, which is wrong. I hope this helps you, English is not my first language, so I don't know if I explained it well enough!

  • abdullahakals

    11 years agoPosted 11 years ago. Direct link to abdullahakals's post “hi this is question i am ...”

    hi this is question i am struggling with any tips or help would be appreciated.
    log_10(log_10 x) -log_10(log_10 3) = log_10 (2)
    this example states simplify and hence solve for X ,

    (8 votes)

    • Wrath Of Academy

      10 years agoPosted 10 years ago. Direct link to Wrath Of Academy's post “We start with `log( log(x...”

      Using the properties of logarithms: multiple steps (video) | Khan Academy (14)

      We start with log( log(x) ) - log( log(3) ) = log(2). Well, first you can use the property from this video to convert the left side, to get log( log(x) / log(3) ) = log(2). Then replace both side with 10 raised to the power of each side, to get log(x)/log(3) = 2. Then multiply through by log(3) to get log(x) = 2*log(3). Then use the multiplication property from the prior video to convert the right side to get log(x) = log(3^2). Then replace both sides with 10 raised to the power of each side again, to get x = 3^2 = 9. And we are done. (And plugging that back in to the original equation with a calculator, it checks out.)

      Yo dog, we heard you like logs, so we put logs in your logs so you can log while you log.

      (20 votes)

  • David Windsor

    10 years agoPosted 10 years ago. Direct link to David Windsor's post “If log (a+c) = Log (b+c) ...”

    If log (a+c) = Log (b+c) does it follow that a+c = b+c ?

    (7 votes)

    • Just Keith

      10 years agoPosted 10 years ago. Direct link to Just Keith's post “Yes, of course. More spec...”

      Using the properties of logarithms: multiple steps (video) | Khan Academy (18)

      Yes, of course. More specifically, it follows that a = b

      (13 votes)

  • Abraham George

    11 years agoPosted 11 years ago. Direct link to Abraham George's post “Who thought of exponents ...”

    Who thought of exponents and logarithms?

    • MrSmall

      11 years agoPosted 11 years ago. Direct link to MrSmall's post “John Napier (a Scottish m...”

      Using the properties of logarithms: multiple steps (video) | Khan Academy (22)

      John Napier (a Scottish mathematician) and Joost Burgi (Swiss mathematician) both discovered them independently around 1600. Napier published "Description of the Marvelous Canon of Logarithms" in 1614, and Burgi published the first lookup tables in 1620.
      Source: The Story of Mathematics, A Rooney, p40

      (11 votes)

  • yanixy12

    12 years agoPosted 12 years ago. Direct link to yanixy12's post “Why is it 2x i got kind o...”

    Why is it 2x i got kind of confused sorry?

    (6 votes)

    • Jason Hansen

      12 years agoPosted 12 years ago. Direct link to Jason Hansen's post “Log_5 25 = 2. So,x*Log_5 ...”

      Using the properties of logarithms: multiple steps (video) | Khan Academy (26)

      Log_5 25 = 2. So,x*Log_5 25 = x*2 = 2x

      (10 votes)

  • Lindsey Fletcher

    10 years agoPosted 10 years ago. Direct link to Lindsey Fletcher's post “In my math textbook the a...”

    In my math textbook the answer to a logarithm is given as log_a(sqrt of a/x)
    I reasoned that this would equal log_a(sqrt of a) - log_a(x)
    The square root of a is equal to a to the 1/2 power. So log_a(a^1/2)=1/2
    Therefore, it simplifies to 1/2- log_a(x)
    Is this wrong or would the book answer be considered simpler?

    (3 votes)

    • Just Keith

      10 years agoPosted 10 years ago. Direct link to Just Keith's post “Your version is correct a...”

      Your version is correct and equal to what the textbook said.

      You are now moving into a level of math where "simplest form" is not going to be as important as whatever might be a useful form or a convenient form. In fact, some values can be expressed in such a variety of ways that even deciding what "simplest form" might be is subjective.

      (7 votes)

  • Daniela Magiricu

    11 years agoPosted 11 years ago. Direct link to Daniela Magiricu's post “How do you find the value...”

    How do you find the value of a logarithm if the number is to the power of log? for example, a^8log base a of square root 2 ?

    (3 votes)

    • Just Keith

      11 years agoPosted 11 years ago. Direct link to Just Keith's post “If the base of both the l...”

      If the base of both the log and the exponent are the same, they cancel each other out. So loga(a^anything) = anything
      It works in the other direction too, so
      a^(loga(anything) = anything
      As long as "anything" is not 0.
      So, in your case you get this:
      a^8loga(√2)
      = a^loga(√2)^8
      (√2)^8 = 2^(8/2) = 2^4 = 16
      Let me know it you didn't follow any of that.

      (6 votes)

  • Clarice

    4 years agoPosted 4 years ago. Direct link to Clarice's post “How will you solve 'log t...”

    How will you solve 'log to base 2 times x = log to base x times 2'?

    (4 votes)

  • Jairus Hasta

    10 years agoPosted 10 years ago. Direct link to Jairus Hasta's post “I'm just confuse how my t...”

    I'm just confuse how my teacher arrive on this solution.... So the problem is
    3^(x-6) = 22
    And he did this
    Log of 3 (22) = x-6
    (Log of 3 (22)) -6=x
    X=-3.19

    (3 votes)

    • Just Keith

      10 years agoPosted 10 years ago. Direct link to Just Keith's post “If you presented your tea...”

      If you presented your teacher's solution correctly, then your teacher is wrong. Specifically, there is a sign error in dealing with the 6.

      Here is one correct way to find the real solution (there are also nonreal solutions, but that is beyond the scope of this level of study).
      log₃[3^(x-6)] = log₃ (22)
      (x - 6) log₃ 3 = log₃ (22)
      (x - 6) = log₃ (22)
      x = 6 + log₃ (22)
      x≈ 8.81359
      Note: if you don't have a calculator that can handle alternative bases of logs, you can use the change of base property: log₃ (22) = ln (22)/ln(3)

      (4 votes)

Using the properties of logarithms: multiple steps (video) | Khan Academy (2024)
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