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Solving exponential equations using exponent properties (advanced)

Sal solves equations like 32^(x/3) = 8^(x-12) and 5^(4x+3) / 25^(9-x) = 5^(2x+5).

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  • blobby green style avatar for user Tessa Pullar
    e^x^2 = -1
    Can you help me solve this..?
    (3 votes)
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    • aqualine tree style avatar for user Judith Gibson
      I think I've solved this, but am not sure.
      e^x^2 = -1
      Write -1 as a vector in polar form
      -1 = cos(3pi/2) + i* sin(3pi/2)
      Use Euler's formula to re-write the polar form
      -1 = e^(3pi/2)*i
      Therefore, e^x^2 = e^(3pi/2)*i
      So, x^2 = 3pi/2 * i
      And x = sqrt (3pi/2 * i)
      Can anyone else agree or disagree this answer?
      (12 votes)
  • leaf blue style avatar for user Jacobi Jackson
    When I had the equation 5x/3=3x-36, I added 36 to both sides which yielded 36+5x/3= 3x. Next, I multiplied both sides by 3 (denominator of 5x/3) and got 36+5x=9x. That gave me a solution of x=9, but I don't see my error.
    (1 vote)
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  • blobby green style avatar for user ItsjustaprankbroXDD
    What if your base, is a fraction, such as 1/2^x-3?
    (4 votes)
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  • aqualine ultimate style avatar for user HanlearnsMath
    At the end of the Second question he has 5^4x+3 / 5^18-2x = 5^2x+5
    But Im very confused as to why he abandoned this ^ up there (the 5 being divided).
    hes acting like 5/5 = 5. that doesnt seem correct.
    (5 votes)
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  • blobby green style avatar for user Deborah Nangereke
    4/3^(x-2) = 3/4^(x-4)
    Solve for x.

    A little stumped. Any help us appreciated.
    (3 votes)
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    • hopper cool style avatar for user Timothy_Lavrenov
      At the start, the equation is:
      4/3^(x-2)=3/4^(x-4)
      To simply, we would multiply both sides by 3^(x-2) and 4^(x-4).
      4*3^(x-2)*4^(x-4)/3^(x-2)=3*3^(x-2)*4^(x-4) /4^(x-4)
      which simplifies to:
      4*4^(x-4)=3*3^(x-2)
      All digits are raised by an invisible one power, so we will add it in:
      4^1*4^(x-4)=3^1*3^(x-2)
      Then you would add the powers:
      4^(x-3)=3^(x-1)
      If f(x)=g(x) then In(f(x))=In(g(x))
      In(4^(x-3))=In(3^(x-1))
      Then we apply the log rule:
      (x-3)In(4)=(x-1)In(3)
      Then we distribute both sides of the equation:
      In(4)x-3In(4)=In(3)x-In(3)
      Then we simplify the equation:
      In(4)x-In(3)x-3In(4)+3In(4)=In(3)x-In(3)x-In(3)+3In(4)
      Which equals to
      In(4)x-In(3)x=-In(3)+3In(4)
      Then we factor out the x
      x(In(4)-In(3))=-In(3)+3In(4)
      Then to make x alone, we divide both sides by In(4)-In(3), and we get
      x=(-In(3)+3In(4))/(In(4)-In(3))
      and we are done.
      (2 votes)
  • aqualine ultimate style avatar for user c_nicole
    Is there a video or lesson for solving exponential equations using factoring? For example: 7(4^2x )=28(4^x )
    (2 votes)
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    • stelly blue style avatar for user Kim Seidel
      I think this is the video that applies the best. The trick with your equation is to recognize that you can divide both sides by 7. You get:
      4^2x =4(4^x ) or 4^2x =4^(x+1 )
      Now, you can apply the concepts in the video to get: 2x = x+1 to solve for "x".
      Hope this helps.
      (4 votes)
  • male robot hal style avatar for user Jonah Chromy
    Why are the equations in the exercise blank?
    (1 vote)
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  • piceratops seed style avatar for user Mansura Hasan
    what is m if 3 to the power of 2m-1 times 9 to the power of m equals to 81.
    In other terms, what is m if; 3^2m-1 x 9^m= 81
    (1 vote)
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    • stelly blue style avatar for user Kim Seidel
      Start by factoring the base values to get a common base for the exponents.
      9=3^2. Thus, 9^m = (3^2)^m = 3^(2m)
      81=3^4

      Now your equation can be written as: 3^(2m-1)*3^(2m)=3^4

      The left side can be simplified further by using properties of exponents: to multiply values with a common base, we add the exponents.
      3^(2m-1+2m)=3^4
      3^(4m-1)=3^4

      Take log base 3 of both sides and you can eliminate the 3's and work with just exponents: 4m-1=4
      Now, you can solve for "m".
      (4 votes)
  • starky tree style avatar for user Chyanne
    At when he rewrites the Left side why doesn't he multiply the 5 and 3 when distributing? Instead he just makes it 5x over 3.
    (1 vote)
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  • blobby green style avatar for user yaheem59
    Hey can I get help with
    3(3^x)-5x(3^x)+2(x^2)(3x)=0. I have to solve for X.
    (2 votes)
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Video transcript

- [Voiceover] So, let's get even more practice solving some exponential equations, and I have two different exponential equations here. And like always, pause the video and see if you can solve for x in both of them. All right, let's tackle this one in purple first. And you might first notice that on both sides of the equation I have different bases. So, it would be nice to have a common base. And when you look at it, you're like, "Well, 32 is not a power of eight, "or at least it's not an integer power of eight." But, they are both powers of two. 32 is the same thing as two to the fifth power, two to the fifth power, and eight is the same thing as two to the third power, two to the third power. So, I can rewrite our original equation, as, instead of writing 32, I could write it as two to the fifth, and then that's going to be raised to the x over three power, X over three power, is equal to, instead of writing eight, I could write two to the third power, two to the third power, and I'm raising that to the x minus 12, x minus 12. Now if I raise something to a power, and then raise that to a power, I could just multiply these exponents. So, I can rewrite the left-hand side as two to the five-x over three, five-x over three power. I just multiply these exponents. And that's going to be equal to two to the... And now I just multiply the three times x minus 12. So, two to the three-x minus 36, and now things have simplified nicely. I have two to this power is equal to two to that power. So, these two exponents must be equal to each other. Five-x over three must be equal to three-x minus 36. So, let's set them equal to each other and solve for x. So five-x over three is equal to three-x minus 36. Let's see, we could, we could multiply, we could multiply everything by three. Let's do that. So if we multiply everything times three, here we're going to get five-x is equal to nine-x minus... What is this? Nine-x minus 108. And now we can subtract nine-x from both sides, and so we will get five-x minus nine-x is gonna be negative four-x is equal to negative 108. We're in the homestretch here. Divide, whoops, sorry about that. We could divide both sides by negative four. Negative four. And we are left with x is equal to, what is this going to be, 27. X is equal to 27. And we are all done. We're all done. And if you had substituted x back in there, you would get 32 to the 27 divided by 3, so 32 to the ninth power, is the same thing as eight to the 27 minus 12th power. So eight to the 15th. Yeah, 27 minus 12, eight to the 15th power. So, anyway, that was fun. Let's do the next one now. So this one looks interesting in other ways. We have rational expressions. We have an exponential up here, exponential down here. And the key realization here is... Well, the first thing I'd like to do. Let me write this 25 in terms of five. We know that 25 is the same thing as five-squared. So, we can rewrite this as five to the four-x plus three over, instead of 25, I could rewrite that as five-squared, and then I'm gonna raise that to the nine minus x, to the nine minus x, and that, of course, is going to be equal to five to the two-x plus five. Now five to the second, and then that to the nine minus x. I can just multiply these exponents. So, this is going to be five to the four-x plus three over five to the... Two times nine is 18, two times negative-x is negative two-x, and that is going to be equal to, that is going to be equal to five to the two-x plus five. And now, let's see. There's multiple ways that we could tackle it. We could multiply both sides of this equation by five to the 18 minus two-x. That's one way to do it. Or we could say, "Hey, look I have five "to some exponent divided by five "to some other exponent, so I could just subtract "this blue exponent from this yellow one." So, the left-hand side will simplify to five to the four-x plus three minus, let me just do minus in a neutral color, minus 18 minus two-x. 18 minus two-x. and that, of course, is going to be equal to what we've had on the right-hand side, five to the two-x plus five. Now we just have to simplify a little bit. Let's see, this is going to be... In fact, we could just say, "Look..." Now I'm having trouble with my little pen tool. Whoops. All right. So now we could say this exponent needs to be equal to that exponent because we have the same base. And so, what we have here on the left-hand side, that I could rewrite as four-x plus three minus 18 plus two-x. I'm just multiplying the negative times both of these terms. So, plus two-x, is going to be equal to two-x plus five. Two-x plus five. So, there's a bunch of different things we could do here. One, we could subtract two-x from both sides, that'll clean it up a little bit. Two-x from both sides. We could also subtract five from both sides. So, let's just do that. So, well, let me just subtract it. Subtract five from both sides. I'm skipping some steps here, but I figure you're at this point reasonably comfortable with linear equations. So then, on the left-hand side, we are going to have four-x, and then you have three minus 18 minus five. Three minus 18 is negative 15, minus five, is negative 20, Is going to be equal to zero. And then, because those cancel out. And so add 20 to both sides, you get four-x is equal to 20. Divide both sides by four, and we get x is equal to five. And we are all done.