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Algebra (all content)

Course: Algebra (all content)>Unit 11

Lesson 16: Solving exponential equations using properties of exponents

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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• e^x^2 = -1
Can you help me solve this..?
• 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?
• 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.
• What if your base, is a fraction, such as 1/2^x-3?
• 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
• 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.
• The rule for dividing numbers with the same base says to write the base once and then subtract the exponents.
• 4/3^(x-2) = 3/4^(x-4)
Solve for x.

A little stumped. Any help us appreciated.
• 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.
• Is there a video or lesson for solving exponential equations using factoring? For example: 7(4^2x )=28(4^x )
• 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.
• Why are the equations in the exercise blank?
(1 vote)
• Maybe you should reload the page.
• 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?
• 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".