Please do more of these "how to solve" type of videos and I hope it becomes a series, it was really easy to follow

@bprpmathbasics

2 ай бұрын

I plan to. Glad to hear they are helpful! Thanks.

@kingsan3162

17 күн бұрын

@@bprpmathbasics Hello,thank you so much and i consider you a teacher of mine even though i never met you live. But i wanna ask you if you can do 2 examples of 1 equation. Becuase in that case it helps anyone see the bigger picture.

12:20 I love how he shows these random bloopers

@lotaniq4449

2 ай бұрын

its an accident lol

@TerraBlo

2 ай бұрын

@@lotaniq4449 he does it quite a lot

@TerraBlo

2 ай бұрын

@@lotaniq4449 he did mean it he puts it on some other videos sometimes

12:19 lazy editing😂 love you bro

-okay, number 6-

@nicholaslama6670

2 ай бұрын

Bro, I got so confused, I thought I was having some recurring déjà vu's

@idontcareforthis

2 ай бұрын

@@nicholaslama6670No déjà-vu for me, but I was on the verge of cursing at my phone for glitching.

@charg1nmalaz0r51

Ай бұрын

@@nicholaslama6670 got stuck in the worst version of groundhog day 🤣

I thought, I have I Internet problem with 'Okay, number 6 what's e though'

@bprpmathbasics

2 ай бұрын

😂

for question 6 I used exponential properties before logarithmic ones

11:15 you can also use ln(pos)/ln(b). most calculators with have natural log in addition to log base 10.

@ronaldjensen2948

2 ай бұрын

For me this was more intuitive. we just go from x ln(5) = ln(2) to x = ln(2)/ln(5)

@Erlewyn

2 ай бұрын

That's actually how I was taught 20+ years ago, in high school in France. I don't think you even learn about different bases log before college here.

excellent job and done at the right pace. I like your work.

Great presentation! Thanks

You take the stress out of math, man. Thank you!

Fantastic review! 🙏🏽👏🏽

You are teaching a lot of people some great maths skills .

Can you make more ''How to solve' videos? I think these videos are easy to follow through, and I want more so I can learn bits of precalc and calc.

Thanks for everything

@penguin9257

2 ай бұрын

This sounds concerning

Exponential equations are so easy that I have never saw on my college and I can solve all of them!!!

It would be nice if people could download a questions sheet of the shown equations but with the numbers changed. To think at home and rewatch your explanation to complete the question. Make a +2 a +5. Something like that for the video questions. The next video give the answers as a 2 sec show frame before the new vid. Mayby win something, Or on patreon or something.

Where is the 25 find the range of function questions? You said you would upload that in your 25 trig questions video.

Most of what you talk about goes over my head, but occasionally I learn something.

Question 7: 3^(x-2) = 5^(x+4) (3/5)^x = 5^4 * 3^2 x = log(5^4 * 3^2)/log(3/5) Question 8: 7^(2x-1) = 2^(4x+3) (7/4)^(2x) = 2^3 * 7 2x = log(2^3 * 7)/log(7/4)

Before watching the video: Question 1: 2↑(3x + 1) = 32 We know that 32 = 2⁵ 2↑(3x + 1) = 2⁵ Comparing like terms: 3x + 1 = 5 3x = 4 x = 0.75. Question 2: 27↑x = 9↑(2x - 3) 27 = 3³ and 9 = 3²: 3↑3x = 3↑(4x - 6) Comparing like: 3x = 4x - 6 x = 6. Question 3: 5↑(x² + 3x - 4) = 1 But 5⁰ = 1: 5↑(x² + 3x - 4) = 5⁰ Comparing like: x² + 3x - 4 = 0 x² + 4x - x - 4 = 0 (x + 4)(x - 1) = 0 x = -4, x = 1. Question 4: (√2)↑(x + 4) = ⅛ But ⅛ = 8↑(-1), 8 = 2² and √2 = 2↑½: 2↑(½x + 2) = 2↑(-3) Comparing like: ½x + 2 = - 3 ½x = -5 x = -10. Question 5: 5↑x = 2 xln5 = ln2 x = ln2/ln5. Question 6: 20e↑(3x - 2) = 1200 e↑(3x - 2) = 60 3x - 2 = ln(60) x = ⅓(ln(60) + 2). Question 7: 3↑(x - 2) = 5↑(x + 4) (x - 2)ln3 = (x + 4)ln5 xln3 - 2ln3 = xln5 + 4ln5 x(ln3 - ln5) = 2ln3 + 4ln5 x = (2ln(3) + 4ln(5))/(ln(3) - ln(5)). Question 8: 7↑(2x - 1) = 2↑(4x + 3) (2x - 1)ln7 = (4x + 3)ln2 2xln7 - ln7 = 4xln2 + 3ln2 x(2ln7 - 4ln2) = (3ln2 + ln7) x = (3ln2 + ln7)/(2ln7 - 4ln2). Question 9: e↑2x + e↑x - 6 = 0: Setting e↑x = t → e↑2x = t²: t² + t - 6 = 0 t² + 3t - 2t - 6 = 0 (t + 3)(t - 2) = 0 t = -3 or t = 2 But as t = e↑x; t > 0: t = 2 → e↑x = 2 x = ln2. Question 10: 2↑x + 3·2↑(-x) = 4 Setting 2↑x = t → 2↑(-x) = 1/t: t + 3/t = 4 t - 4t + 3 = 0 t - 3t - t + 3 = 0 (t - 1)(t - 3) = 0 t = 1 or t = 3. But t = 2↑x: Case 1: t = 1: 2↑x = 1 x = 0. Case 2: t = 3: 2↑x = 3 xln2 = ln3 x = ln(3)/ln(2).

@Apollorion

2 ай бұрын

I think you made a pair of typo's in Question 10; after multiplying the whole equation with t, t does not turn into t² whereas it should and is also processed as.

@GirishManjunathMusic

2 ай бұрын

@@Apollorion so I have! The radical must not have entered properly, I'm typing these on the phone keyboard, so those radicals are sometimes difficult to enter!

@Apollorion

2 ай бұрын

@@GirishManjunathMusicWell, bless KZread. /s If I made a 'mistake' from the perspective of the 'algorithm' then I want to know how/why and not a brutal disappearance without explanation; I want to learn from my mistakes to avoid to repeat them. P.S. b.t.w. you might help yourself with copy & paste or describing t squared with other symbols e.g. t^2 or tt.

@thewendigo9263

Ай бұрын

You made a mistake on question one. X is 1.33

@Apollorion

Ай бұрын

@@thewendigo9263I agree with the mistake you seem to notice (i.e. 3x=4 => x=4/3) but not with the alternate solution you propose i.e. 1.33 _It is more accurate but still not correct_ . The correct solution in expressed in decimals would finish with an infinite amount of 3s after the dot, not just two.

Love the video. But I saw you miss this a couple of times now. The reason the exponential property of same bases works is because if you take the log of both sides they cancel. _Note I saw you covered this later, but it would be helpful to explain in problem 1._ 2^(3X+1) = 2^5 log₂ 2^(3X+1) = log₂ 2^5 (how do I do strikeout in a comment?) Which leaves you with ... 3X+1 = 5 Heads up - that extra explanation will make things clearer for some students. Great job though. I love your videos!

@cheeseplated

2 ай бұрын

-for strikethrough- put - on either end of the text _for italics_ it's _ *for bold* it's * *_-and you can combine them-_*

@bprpmathbasics

2 ай бұрын

Thank you!

@Apollorion

2 ай бұрын

The reason that that 'exponential property of same bases' works is that: A: the base is bigger than 0 as well as well unequal to 1 and B: if y=a^x and z=dy/dx then z is unequal to 0 and has a constant sign (i.e. d( z/|z| )/dx = 0 no matter the value of x) which implies that for an increase in x, y is always bigger (for a>1) or smaller (for a

i am promoted to class 10th 2 months ago and i solved all. I cant believe it.

@4:46 OMG, I know that 5^0 = 1. How did I not see that I needed to use that when I tried the problem before I watched you solve it?????

@darranrowe174

2 ай бұрын

Sometimes these facts are so simple, we miss them. When I was younger, I know I often saw equality as a one way thing so I could apply it one way but never thought about applying it the other. This is why videos like this can be good. It makes you think of things in ways you never thought of before.

@fullc0de

2 ай бұрын

@@darranrowe174 When you do make the connection and see how the puzzle piece fits, it's such a great feeling, no matter how simple the insight is! When you don't, it's a total Homer Simpson '"D'oh!" moment! The drama of math.

What do you do when the teacher tells you to express the answer to #7 and #8 as a single logarithmic function? 😂

@bprpmathbasics

2 ай бұрын

Show them this video: kzread.info/dash/bejne/iYp7vMSIYMjZf7A.html : )

Can you solve log(x)=x?

@cdkw8254

2 ай бұрын

e^x

@cpu_1292

2 ай бұрын

You raise x by both sides to get x^log(x) = x^x The log and the x cancels out so you get x = x^x x has to be one because 1 = 1^1 Edit: I made a mistake, there is no real solution Edit2: here is the real solution Raise 10 by both sides and get 10^log(x) = 10^x The 10 and the log cancel out to get x = 10^x so there is no real solution

@smokeless-flame2025

2 ай бұрын

@@cpu_1292 i think thats wrong bc log(1)=0 [10^0=1]

@smokeless-flame2025

2 ай бұрын

no real solutions

@xinpingdonohoe3978

2 ай бұрын

Not in the real plane. This can be shown by knowing ln(x)=x is the same as x=e^x. x2>1 means e^x>2^x, so e^x>x, so x>ln(x).

5:16 i factor (x-3) (x-1) am i wrong?

@zilovesmaths

Ай бұрын

Ah you've done it the other way around To factor it you need 2 numbers that multiply to give c (-4) And they need to add to give b (+3) You've got 2 numbers that multiply for b and add for c haha So if you expand (x-3)(x-1) That gives x²-4x+3 instead of x²+3x-4

Can you please upload more on your main channel? I miss the exciting content on that channel

i love you brro

~I Found All Of Them Ez

and as always that's it lol.