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Using the logarithmic power rule

Sal rewrites log₅(x³) as 3log₅(x). Created by Sal Khan and Monterey Institute for Technology and Education.

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  • blobby green style avatar for user John Emmett
    How do you do log2 -1
    (23 votes)
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  • orange juice squid orange style avatar for user Lasha
    what's log2(2x) - log2x^3 = 5 ?
    (6 votes)
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  • blobby green style avatar for user Fin Teasley
    I'd like to see a step by step solution to the following equation: 5^(3x+2)=3^(4x-1) solve for x
    (9 votes)
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  • mr pants teal style avatar for user Cooper Young
    What happens if you have logarithms involving imaginary or complex numbers?
    (6 votes)
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  • old spice man green style avatar for user doug
    at about , the instructor states that (a^b)^d=a^bd. How is it possible for the exponent to shift down a degree?
    (3 votes)
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    • male robot hal style avatar for user Lars Opstad
      This is how exponents work. It doesn't shift it down a degree. Here are a few cases and examples:

      (2^3)^2=8^2=64
      (2^3)^2=2^(3*2)=2^6=64 (2^1=2,2^2=2,2^3=8,2^4=16,2^5=32,2^6=64)

      (3^2)^3=9^3=729
      (3^2)^3=3^(2*3)=3^6=729 (3,9,27,81,243,729)

      This is not unlike what happens when you multiply two numbers with the same base raise to different powers. In that case you add the exponents:

      2^2*2^4=2^(2+4)=2^6=64
      2^2*2^4=4*16=64

      Good question. It is confusing at first, but when you get the hang of it, it's not too hard.
      (6 votes)
  • piceratops seedling style avatar for user Fredo
    Having a problem what is log(1000)= 4.605? This problem doesn’t seem to make sense to me and many of the problems following are similar
    (2 votes)
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  • aqualine ultimate style avatar for user RezLeo
    And yet another doubt I stumbled upon while exercising:
    Give the domain of
    f(x)=log(base3)(4x-3)^2

    First solution:
    2*log(base3)(4x-3)
    4x-3>0
    x>3/4

    Second solution:
    log(base3)(4x-3)^2
    (4x-3)^2>0
    x different than 3/4

    Why is the first solution wrong?
    (2 votes)
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    • piceratops ultimate style avatar for user Just Keith
      The domain depends on the positioning of the square, but your notation is ambiguous.
      If you meant: f(x) = log₃[(4x-3)²]
      Then the domain is x≠ ¾ because that value would lead to a log of 0. The square makes sure we never get a negative value for the argument.

      If you mean f(x) = {log₃[(4x-3)}² which is usually written as log²₃ (4x-3) then:
      As before x≠ ¾ due to being a log of 0. However, you are squaring the entire log, not just the argument. Thus, the argument can be negative, and the log of a negative number is not real, so that is outside the domain. Thus, the domain is
      x > ¾

      Again, please remember, the property log (aⁿ) = n log (a) ONLY APPLIES if a > 0. Therefore, you shouldn't use this property with variables that might lead to the argument being negative because that may lead to extraneous solutions.
      (2 votes)
  • leafers seedling style avatar for user Cristina  Moreno
    How would you solve 6 log x=-3 ?
    (2 votes)
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    • old spice man green style avatar for user Petrie (Peter S. Asiain III)
      Okay I'll show you the solution.
      6log(x) = -3 →This the problem we are given
      log(x) = -3/6=-1/2 → I divided both sides by 6
      10^log(x) = 10^(-1/2) →Raised both sides by base 10
      x = 10^(-1/2) → Since the base of log is 10, 10^log(x)=x, and we are done. We now know the value of x.

      Take note that log(x ) = log10(x). If you see a log without a specified base, you can automatically assume that its base is 10.
      (2 votes)
  • male robot johnny style avatar for user Alexander Yellen
    How would you do something such as: solve 2^(x^2)=8 for x?
    (1 vote)
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  • piceratops seed style avatar for user tirkeyajaystar1999
    If a= log12 base 24 , b= log24 base36 , c=log36 base 48 , then value of (1+abc)/bc equals to ?
    (1 vote)
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    • aqualine tree style avatar for user Judith Gibson
      I think that the answer is 2. Here's why.
      Let a = log_24 (12) , b = log_36 (24), c = log_48 (36)
      First, since neither b or c are zero, I would simplify ( 1 + abc ) / bc to ( 1 / bc) + a.
      = 1 / (log_36 (24 ) * log_48 (36) ] + log_24 (12)
      Now use the change of base rule to get base 10 logs for everything.
      1 / [ ( log24 / log36 ) * ( log36 / log48 ) ] + log12 / log24
      = ( log48 / log24 ) + ( log12 / log24 )
      = ( log48 + log12 ) / log24
      = log (48*12) / log24
      = log (24*24) / log24
      = ( log24 + log24 ) / log24
      = 1 + 1
      = 2
      Any questions?
      (3 votes)

Video transcript

We're asked to simplify log base 5 of x to the third. And once again, we're just going to rewrite this in a different way. You could argue whether it's going to be more simple or not. And the logarithm property that I'm guessing that we should use for this example right here is the property-- if I take log base x of-- let me pick some more letters here, log base x of y to the zth power. This is the same thing as z times log base x of y. So this is a logarithm property. If I'm taking the logarithm of a given base of something to a power, I could take that power out front and multiply that times the log of the base, of just the y in this case. So we apply this property over here. And in a second, once I do this problem, we'll talk about why this actually makes a lot of sense and comes straight out of exponent properties. But if we just apply that over here, we get log base 5 of x to the third. Well, this is the exponent right over here. That's the same thing as z. So that's going to be the same thing as-- let me do this in a different color-- that 3 is the same thing-- we could put it out front-- that's the same thing as 3 times the logarithm base 5 of x. And we're done. This is just another way of writing it using this property. And so you could argue that this is a what-- maybe this is a simplification because you took the exponent outside of the logarithm, and you're multiplying the logarithm by that number now. Now with that out of the way, let's think about why that actually makes sense. So let's say that we know that-- and I'll just pick some arbitrary letters here-- let's say that we know that a to the b power is equal to c. And so if we know that-- that's written as an exponential equation. If we wanted to write the same truth as a logarithmic equation, we would say logarithm base a of c is equal to b. To what power do I have to raise a to get c? I raise it to the bth power. a to the b power is equal to c. Fair enough. Now let's take both sides of this equation right over here, and raise it to the dth power. So let me make it-- so let's raise-- take both sides of this equation and raise it to the dth power. Instead of doing it in place, I'm just going to rewrite it over here. So I wrote the original equation, a to the b is equal to c, which is just rewriting this statement. But let me take both sides of this to the dth power. And I should be consistent. I'll use all capital letters. So this should be a B. Actually, let's say I'm using all lowercase letters. This is a lowercase c. So let me write it this way, a to the-- so I'm going to raise this to the dth power, and I'm going to raise this to the dth power. Obviously, if these two things are equal to each other, if I raise both sides to the same power, the equality is still going to hold. Now, what's interesting over here is we can now say-- what we could do is we can use what we know about exponent properties. Say, look, if I have a to the b power, and then I raise that to the d power, our exponent properties say that this is the same thing. This is equal to a to the bd power. This is equal to a to the bd. Let me write it here. This is-- let me do that in a different color. I've already used that green. This right over here, using what we know about exponent properties, this is the same thing as a to the bd power. So we have a to the bd power is equal to c to the dth power. And now this exponential equation, if we would write it as a logarithmic equation, we would say log base a of c to the dth power is equal to bd. What power do I have to raise a to get to c to the dth power? To get to this? I have to raise it to the bd power. But what do we know that b is? We already know that b is this thing right over here. So if we substitute this in for b, and we can rewrite this as db, we get logarithm base a of c to the dth power is equal to bd, or you could also call that db, if you switch the order. And so that's equal to d times b. b is just log base a of c. So there you have it. We just derived this property. Log base a of c to the dth, that's the same thing as d times log base a of c, which we applied right over here.