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## Integrated math 2

### Course: Integrated math 2 > Unit 6

Lesson 5: Solving exponential expressions using properties of exponents# Solving exponential equations using exponent properties (advanced)

CCSS.Math: , ,

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

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- e^x^2 = -1

Can you help me solve this..?(3 votes)- 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?(13 votes)

- 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)
- Your error is in your multiplication step. If you multiply the entire equation by 3, you need to do:

36(3) + 5x/3 (3) = 3x (3)

You skipped the 36 and only multiplied the other 2 terms by 3.

Hope this helps.(13 votes)

- What if your base, is a fraction, such as 1/2^x-3?(4 votes)
- A fraction means a negative exponent, so what you would do to invert it is bring the denominator up to the numerator and make the exponent negative(7 votes)

- 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)- The rule for dividing numbers with the same base says to write the base once and then subtract the exponents.(4 votes)

- 4/3^(x-2) = 3/4^(x-4)

Solve for x.

A little stumped. Any help us appreciated.(3 votes)- 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.(3 votes)

- Is there a video or lesson for solving exponential equations using factoring? For example: 7(4^2x )=28(4^x )(2 votes)
- 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)

- Why are the equations in the exercise blank?(1 vote)
- Maybe you should reload the page.(3 votes)

- on the practice there was a question

1.) 49^(-x-7) * 7^(4x+8) = 49^(5x-3)

2.) 7^(-2x-14) * 7^(4x+8) = 7^(10x-6)

3.) 7^(2x-6)=7^(10x-6)

4.) 2x-6=10x-6

5.) 2x=10x ? pls tell me where i made a mistake, or should i just type 0 as the answer?(2 votes)- If you plug 0 back into the original equation for "x" does it make both side equal? If it does, then you have a valid solution. I did it quickly in my head and I think it's ok. But, check if yourself to confirm.(2 votes)

**QUESTION**

Can someone give me a brief overview of this video with an equation example to help me understand this concept better?

Thankyou and have a great day! 🙂(1 vote)- Lets go backwards from a solution to a problem to see if it helps which will also help you create your own problems and hopefully gives you a better understanding of the concept. So let x = 4. Next, we we create two expressions using distribution, start with one side 2(2x+5) for x=4, you get 2(8+5)=26, but if I want something divisible by 3 (so I can create a distribution with 3 on outside). I change it to 2(2x+7) for x=4 gives 2(8+7)=30. So for x=4, lets create a 3(4x-6). I started with 30, divided by 3 to get 10, then put 4*4=16 and subtracted 6 to get 10, so we have 2(2x+7)=3(4x-6). Now I add some base to it, just choose 3. This gives 3^(2(2x+7) = 3^(3(4x-6). Then just do the first two parts, 9^(2x+7) = 27^(4x-6). So to solve, we would just go backwards step by step. Does this help give a better understanding?(4 votes)

- 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)- 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)

## 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.