@@mshavrot_math I think I’m still not understanding a step. even if I separate the two exponents, I keep ending up with the wrong answer. What happens to the 3^(6^x)?
Ok, I think I know why you are confused. It would have been better had I said .. let a =3^x Then you would get the quadratic a^2 -5a +6 =0 I think you are confused with all of the x’s. At the end substitute back in the 3^x for a.
@ I figured it out Ms, you first factor out x and then do (1-1.585) which equals 0.585x, you did it all in one go that’s why I got confused. Thank you Ms!
Yes. Go to question 12 .... First you need to rewrite 6(25)^x as 6(5)^2x .... then you have 6(5)^2x - 17(5)^x +12 = 0 Now, let 5^x = a Rewrite the equation as 6a^2 -17a +12 = 0 Now factor to get (2a-3)(3a-4) = 0 Remember at this point that you need to sub back in 5^x for a and follow the technique that I used in #12 in this video. Let me know if you figure it out!
Have you started chapter 9 in your class already? I was going to do trig next but if other schools are doing chapter 9 first I will do that first. Let me know.
@@mshavrotscanadianuniversit6234 thank you. my school teacher start chapter 1 with 9 because they side they are similar to each other. but now we done the test and the teacher comeback to chapter 2 now. can you please do more words problems and give some tips to answer quickly the questions. thank you for ever think again
Yes I do! First let x = 2^x That will give you the following 2^x - 1/(2^x) = 4 X - 1/x - 4 = 0. Multiply by x X^2 -1 -4x = 0 X^2 -4x - 1 = 0 Now solve using the quadratic formula and you should get 4.236 or -0.2361. Remember this is for x = 2^x Don’t plug this back into the original equation but into the one where you let x = 2^x and you will get 4.236 - 1/4.236 = 4 which is true. Now that means that 2^x = 4.236 Take the log base2 of each side and you will get x= log 4.236/ log2 And x=2.0827. Tahdah Now you can plug that into the original equation to check.also you should test your second solution of -0.2361 and see what happens
Ms. Havrot, you explain each step so well 👏
Ms Havorits, is the biggest beauty around so thankful for her shes the reason im doing decent in functions and calculus Gr12
So happy to have been able to help you with your studies ❤️ Thanks for watching and for your lovely comment 🥰
Hi Ms Harvot! I got sort of lost at 20:15. how do i substitute the 3^x into x to get x^2 and 5(x)?
You need to recognize that 3^(2x) is the same as (3^x)^2 then when you let x = 3^x you get x^2
Does that help??
@@mshavrot_math I think I’m still not understanding a step. even if I separate the two exponents, I keep ending up with the wrong answer. What happens to the 3^(6^x)?
Ok, I think I know why you are confused. It would have been better had I said .. let a =3^x
Then you would get the quadratic a^2 -5a +6 =0
I think you are confused with all of the x’s. At the end substitute back in the 3^x for a.
@@mshavrot_math Ohh that makes sense! I’m sorry about that, I misunderstood.
and thank you for the help :)
Hi Ms. Harvot, was wondering if I could use the same strategy you used in the last question of the video to solve question 12a) on the unit review?
Yes! Good for you for recognizing the pattern 😊
Thank you for your hard work Ms Havrot🌹🌹all i have is appreciation❤️❤️
So nice of you to say. : )
Hello Ms, I hope you are doing well. At 13:55 how does x - 1.585x = .585x?
It doesn’t. Watch it again as I think you missed something.
@ I figured it out Ms, you first factor out x and then do (1-1.585) which equals 0.585x, you did it all in one go that’s why I got confused. Thank you Ms!
Hi Ms Havort! Do you have any idea on how to solve 6(25)^x - 17(5^x) = -12 ?
Yes. Go to question 12 .... First you need to rewrite 6(25)^x as 6(5)^2x .... then you have 6(5)^2x - 17(5)^x +12 = 0 Now, let 5^x = a Rewrite the equation as 6a^2 -17a +12 = 0
Now factor to get (2a-3)(3a-4) = 0 Remember at this point that you need to sub back in 5^x for a and follow the technique that I used in #12 in this video. Let me know if you figure it out!
@@mshavrotscanadianuniversit6234 That clears things up a lot! Thank you so much! :)
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@@mshavrotscanadianuniversit6234 Of course! I've recommended you to my friends in the past because your channel is so helpful!
Thank you! I appreciate the support 😊
This woman is a goddess. Thank you!
simp alert
@@rasimqusous7715 she helped me pass grade 12 functions so yea, I’m a simp
Remember that I’m not responsible for anyone’s grades so sucking up to me does nothing but show appreciation which is basically what makes my day. 😊
Hi Ms Havort can u check the problem on 20:11 I think it is 2(x)
I’m substituting x in for 3^x so because it started as 3^(2x) that would leave me with x^2. Does that help?
hello your videos is amazing I always watching it
can you please make chapter 9 Adv function and make more word problems
As soon as possible
Have you started chapter 9 in your class already? I was going to do trig next but if other schools are doing chapter 9 first I will do that first. Let me know.
@@mshavrotscanadianuniversit6234
thank you. my school teacher start chapter 1 with 9 because they side they are similar to each other. but now we done the test and the teacher comeback to chapter 2 now. can you please do more words problems and give some tips to answer quickly the questions. thank you for ever think again
Hello Miss, Do you have any idea on how to solve 2^x - 2^-x = 4 ??
Yes I do! First let x = 2^x
That will give you the following
2^x - 1/(2^x) = 4
X - 1/x - 4 = 0. Multiply by x
X^2 -1 -4x = 0
X^2 -4x - 1 = 0
Now solve using the quadratic formula and you should get 4.236 or -0.2361. Remember this is for x = 2^x
Don’t plug this back into the original equation but into the one where you let x = 2^x and you will get
4.236 - 1/4.236 = 4 which is true.
Now that means that 2^x = 4.236
Take the log base2 of each side and you will get x= log 4.236/ log2
And x=2.0827. Tahdah
Now you can plug that into the original equation to check.also you should test your second solution of -0.2361 and see what happens
Thank you so much! @@mshavrotscanadianuniversit6234
ty
Havrot*