Main content

## Algebra 2

### Course: Algebra 2 > Unit 6

Lesson 5: Solving exponential equations using properties of exponents- Solving exponential equations using exponent properties
- Solve exponential equations using exponent properties
- Solving exponential equations using exponent properties (advanced)
- Solve exponential equations using exponent properties (advanced)
- Rational exponents and radicals: FAQ

© 2023 Khan AcademyTerms of usePrivacy PolicyCookie Notice

# Solving exponential equations using exponent properties

CCSS.Math: , ,

Sal solves equations like 26^(9x+5) = 1 and 2^(3x+5) = 64^(x-7).

## Want to join the conversation?

- wait... so in the first equation he divided -5 by 9 and kept -5 instead of making it.5 repeating. why?(25 votes)
- This is a known error in the video. If you watch the video in normal mode (not full screen), you would see a correction box pop-up and tell you that Sal meant x=-5/9(52 votes)

- how would i solve 2^-2b = 2^-b(6 votes)
- For the 2 sides of your equation to be equal, the exponents must be equal.

So, you can change the equation into:`-2b = -b`

Then, solve for "b"

Sal does something very similar at about3:45in the video.

Hope this helps.(15 votes)

- What is the operation called when you turn a fraction into a multiple of some number...? Like
`1/32`

becomes`2^-5`

! And is there any way to perform this sort of alteration easily!?^_^"

Thank you kindly in advance.(5 votes)- You just rewrote the number in its factored form using prime factors and exponents.

If you know how to do prime factorization and division, breaking the number down is not hard.(10 votes)

- I can't believe this... not only do we have normal equations...but we now have equations on the exponents... this is pretty neat, it's like we're "zooming in" on the base number's exponents who have their own "world" of math going on, right above the base numbers that we normally work with. I do have a question though. For the first problem with the 26=1...couldn't you raise the 26 to the zero power...then have that 0(9x+5) multiply out and equal 0...then it would leave 26^0 which is 1...and that would equal 1=1? This isn't how Sal does it, but I think it's a valid method?(5 votes)
- That is a valid operation that results in a true statement, but it doesn't actually help us find x. We can also multiply both sides of an equation by 0, but that's not helpful either.(5 votes)

- he did the first one wrong(4 votes)
- Yes, but there is a correction box in bottom right at around1:05which noted this and corrected it.(4 votes)

- How would you solve

3^(x+3)-3^(x)=78?(3 votes)- I can only get to 3^x = 78/8.(1 vote)

- in the first one isnt x= -5/9?(4 votes)
- Yes. This is a known error in the video. A correction box pops up at about4:55in the video and tells you Sal meant to write x=-5/9.(1 vote)

- At4:08, I don't get why x^a=x^b proves that a=b. Can someone please give me justification on why that is the case?(2 votes)
- Try it with some numbers

if x=2 and a=3, then 2^3=2^b

What exponent can you use for "b" to get the same result as 2^3? The only value that will work is if b=3.

Hope this helps.(4 votes)

- Is this the same thing as equations with exponents or no?(2 votes)
- These are equations where the variable is an exponent. There are other equations like a quadratic equation where the variable has a numeric exponent. Which type are you looking for?(3 votes)

- Before seeing the obvious strategy in the first example is to use the x^0 = 1 I calculate the answer to be x = -5/4. I wonder if somebody could help identify where my logic is incorrect:

26^9x+5 = 26^9x`*`

26^5 (split off +5)

(26^9)^x`*`

26^5 (factor out ^x)

(26^5 * 26^4)^x`*`

26^5 (rewrite 26^9 as product of ^5+4)

26^5(26^4)^x (factor out 26^5*)

26^4x = 1/26^5 (divide by 26^5)

26^4x = 26^-5 (reciprocate the 1/26^5)

4x = -5 (solve for x)

-> x = -5/4 (answer)

*Is it that I factor out incorrectly here? I cant factor out one 26^5 as there is essentially ^x amount of 26^5's so can't reduce it out as just one?

Many Thanks to whichever kind stranger knows!(2 votes)- One problem is is in step 4, you cannot factor out things using multiplication, factoring out is used with adding terms. Lets use simpler terms (2^2*2^3)^3*2^2 = (4*8)^3*4 = 131072. However, 2^2(2^3)^3=2048, so these two are clearly not the same. You would have (26^5)^(x+1)*26^4 which is not getting you any closer to the answer. I think you realized the issue of factoring by your question near the end.(3 votes)

## Video transcript

- [Voiceover] Let's get some practice solving some exponential equations, and we have one right over here. We have 26 to the 9x plus five power equals one. So, pause the video and
see if you can tell me what x is going to be. Well, the key here is to realize that 26 to the zeroth power, to the zeroth power is equal to one. Anything to the zeroth power is going to be equal to one. Zero to the zeroth power we can discuss at some other time, but anything other than zero to the zeroth
power is going to be one. So, we just have to
say, well, 9x plus five needs to be equal to zero. 9x plus five needs to be equal to zero. And this is pretty
straightforward to solve. Subtract five from both sides. And we get 9x is equal to negative five. Divide both sides by nine, and we are left with x is
equal to negative five. Let's do another one of these, and let's make it a little bit more, a little bit more interesting. Let's say we have the exponential equation two to the 3x plus five power is equal to 64 to the x minus seventh power. Once again, pause the video, and see if you can tell
me what x is going to be, or what x needs to be to satisfy
this exponential equation. All right, so you might, at first, say, oh, wait a minute,
maybe 3x plus five needs to be equal to x minus seven,
but that wouldn't work, because these are two different bases. You have two to the 3x plus five power, and then you have 64 to the x minus seven. So, the key here is to
express both of these with the same base, and lucky for us, 64 is a power of two, two to the, let's see, two to the third is eight, so it's going to be two to the
third times two to the third. Eight times eight is 64, so it's two to the sixth is equal to 64, and you can verify that. Take six twos and multiply them together, you're going to get 64. This was just a little bit easier for me. Eight times eight, and
this is the same thing as two to the sixth power, is 64, and I knew it was two to the sixth power because I just added the exponents because I had the same base. All right, so I can rewrite 64. Let me rewrite the whole thing. So, this is two to the 3x plus five power is equal to, instead of writing 64, I'm going to write two to the sixth power, two to the sixth power, and then that to the x minus seventh power, x minus seven power. And to simplify this a little bit, we just have to remind ourselves that, if I raise something to one power, and then I raise that to another power, this is the same thing as raising my base to the product of these powers, a to the bc power. So, this equation I can rewrite as two to the 3x plus five is equal to two to the, and I just multiply six times x minus 7, so it's going to be 6x, 6x minus six times seven is 42. I'll just write the whole thing in yellow. So, 6x minus 42, I just multiplied the six times the entire expression x minus 7. And so now it's interesting. I have two to the 3x plus 5 power has to be equal to two
to the 6x minus 42 power. So these need to be the same exponent. So, 3x plus five needs to
be equal to 6x minus 42. So there we go. It sets up a nice little
linear equation for us. 3x plus five is equal to 6x minus 42. Let's see, we could get all of our, since, I'll put all my Xs
on the right hand side, since I have more Xs on the right already, so let me subtract 3x from both sides. And let me, I want to
get rid of this 42 here, so let's add 42 to both sides, and we are going to be left with five plus 42 is 47, is equal to, 47 is equal to 3x. Now we just divide both sides by three, and we are left with x is
equal to 47 over three. X is equal to 47 over three. And we are done.