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Solving exponential equations using logarithms: base-10

Sal solves the equation 10^(2t-3)=7. Created by Sal Khan.

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  • leaf blue style avatar for user Earl James
    Couldn't we just oversimplify it to (log 7000)/(log 100) ?
    (6 votes)
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    • primosaur seed style avatar for user Ian Pulizzotto
      Yes, this can be done. We can see that (log 7000)/(log 100) is equivalent to the correct answer given, which is [(log 7) + 3]/2, using definitions and laws of logarithms:

      (log 7000)/(log 100) = [log(7*10^3)] / [log (10^2)] = [(log 7) + log (10^3)] / [log (10^2)]
      = [(log 7) + 3]/2.

      Have a blessed, wonderful day!
      (11 votes)
  • blobby green style avatar for user ky thomas
    I am having trouble solving 3log(4x+3) < 1. When I computed the problem I got x > 5/6.
    I 1st divided both sides by 3, then set it up as e^4x + e^3 < e^1/3. which i then rewrote as 4x+3 < 1
    I don't think my answer is correct but I am lost as to where I went wrong.
    Pease help!!
    (2 votes)
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    • primosaur ultimate style avatar for user Mike Runge
      Dividing by 3 is the correct 1st step, but from there, I like to convert it back to its exponential form. That means that if this is a "common logarithm" (base 10), you would write 4x+3 < 10^(1/3). This can also be attained by raising both sides to a power of 10 [i.e. 10^(log(4x+3)) < 10^(1/3)]. The 10^log cancels itself out and you are left with the 4x+3 on the left. Solving 4x+3 < 10^(1/3) leaves us with x < [-3 + 10^(1/3)] / 4 or approximately x < -0.2114. If you plug it into the original inequality, you can verify the correct answer.
      (7 votes)
  • blobby green style avatar for user beccacollin123
    I am having trouble with solving a exponential equation using a logarithm. The equation is 5(10)^x-31=81.6 , I keep trying every way to solve this but it's just not giving me the correct answer. Help!
    (4 votes)
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    • hopper happy style avatar for user Rashel
      1st Isolate the base with the exponent by dividing both sides by 5 and you get:
      10^x-31=16.32

      2nd log both sides

      log 10 of 10^x-31=log 10 of 16.32
      The log 10 and 10 cancel out, your left with:
      x-31=log 10 of 16.32

      3rd add 31 to both sides to isolate x
      x=log 10 of 16.32 +31

      4th Depending on your calculator, you will either press the log button first and then enter the value or you will enter the value first and then press the log button.
      For the windows calculator: Type in 16.32 and then click the log button and then click on enter.
      You get 1.2127
      Then add 31 to that value: 32.213
      ( I rounded to the nearest thousandth)
      Check to see if answer is correct:
      Plug 32.213 into x
      10^32.213=1.63305...
      1.63305*5=8.165259...
      Subtract 31 from that number
      8.165259
      So yes, our answer is correct
      (3 votes)
  • blobby green style avatar for user Bob D'Arcy
    7=x^2.8073 how do you solve this with logs
    (2 votes)
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  • winston baby style avatar for user Shreyaan Jain
    Is there a video for an e-base problem instead of a base with 10?
    (3 votes)
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  • aqualine sapling style avatar for user Sophia Navarre
    What happens if you have a power on either side, for example: 4^2x+3 = 5^3x-1
    (2 votes)
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    • blobby green style avatar for user Natasha
      Take logs on both sides. log(4^2x+3) = log(5^3x-1)
      Then use the log rules to bring down the power 2x+3log4 = 3x-1log5
      You can then split these logs up 2xlog4 + 3log4 = 3xlog5 - log5
      Get all your x values over to one side 3xlog5 - 2xlog4 = 3log4 + log5
      Then factorise to take out x x (3log5 - 2log4) = 3log4 + log 5
      And rearrange to find x x = (3log4 + log5) / (3log5 - 2log4)
      (4 votes)
  • aqualine tree style avatar for user Emilee Weir
    So I'm studying for a test. In one of my practice problems, I have to solve for x in the equation 12^(x-3)=17^(2x), and I am supposed to write the exact answer in terms of log_10. How do I solve it?

    One other thing I'm curious about: why does solving for a variable in terms of log_10 work?
    (3 votes)
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    • aqualine ultimate style avatar for user Johnathan
      Since we're dealing with variables as indices (aka powers, exponents), log_10, log_(anything), and ln (the natural log) all work because they are inverses of exponential functions. By using log, you can easily manipulate the exponents making it easier to solve.
      12^(x - 3) = 17^(2x) {Equation from Question}

      log(12^(x - 3)) = log(17^(2x)) {Log both sides} I'll be referring to log_10 as log from now on.

      (x - 3)log(12) = 2x(log(17)) {Standard rule with logarithms, essential for solving these types of questions}

      xlog(12) - 3log(12) = 2x(log(17)) {Distributing on Left Hand Side}

      xlog(12) - 2x(log(17)) = 3log(12) {Reorganising the Equation}

      x(log(12) - 2log(17)) = 3log(12) {Factoring Out 'x'}

      x = 3log(12)/(log(12) - 2log(17)) {Isolate 'x'}


      And I think I answered your second question on top. Hope this helps. :-)
      (1 vote)
  • piceratops ultimate style avatar for user mrinmoy
    "-(-2)^n = 1532" how do I figure out the value of n?
    (2 votes)
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    • purple pi pink style avatar for user Mariam
      open up the brackets and you will get positive 2.
      now as far my brain works. you can factor out 1532.
      this would be a lengthy task so you can apply a shortcut. we know 2^4=16
      so divide 1532 by 32 and you get let's say x.
      then multiply x by 4 and here you get the answer.
      in case there is a remaider factor out the remainder and add the result to x*4.
      hope it helps.
      (2 votes)
  • piceratops sapling style avatar for user Eliza
    with all the crazy things you can do with a graphing or scientific calculator, why don't any of them have the ability to let you choose your log base to something other than e or 10?
    (2 votes)
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  • blobby green style avatar for user quinzubella
    Why is 10 specifically used as the common logarithm base?
    (1 vote)
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Video transcript

Voiceover:Solve the equation for T and express your answer in terms of base 10 logarithms. And this equation is 10 to the 2T - 3 is equal to 7. We want to solve for T in terms of base 10 logarithms. So let me get my little scratchpad out and I've copied and pasted the same problem. So I'm just going to rewrite it, so they have 10 to the 2T-3 is equal to 7. Actually, let me color code this a little bit. So 10 to the 2T - 3 is equal to 7. So this is clearly an exponential form right over here. if we want to write it in logarithmic form, where we could, that'll essentially allow us to solve for the exponent, so we could say, this is the exact same truth about the universe as saying that the log base 10 of 7 is equal to 2T - 3. Let's just make sure that makes sense, this is saying 10 to the 2T - 3 = 7 This is saying that the power that I need to raise 10 to to get to 7 is 2T - 3, or 10 to 2T - 3 power is equal to 7. So these are equivalent statements. What this form does is it starts to put it into a form that's easier to solve for T, now if we want to solve for T, we can add 3 to both sides. So if we add 3 to both sides, we are going to get, log base 10 of 7 + 3, plus 3. And this +3 is of course outside of the logarithm to make it clear, it's just like that. And on the right hand side this is going to be equal to, this is going to be equal to just 2T. And now to solve for T we just divide both sides by 2. So if we divide both sides by 2 we get, T is equal to all of this business, log base 10 of 7 + 3 all of that over 2, so let me see if I can remember this, and write it in the actual box that they gave us. So we want to do log base 10 of 7 and then that + 3 yup it's formatting it right, divided by 2. So that's what I'm claiming the T is equal to in terms of Base 10 logarithms. So let me check my answer and I got it right.