Fortunately this old man could solve it I love these little challenges ❤

Thank you sir for making this video , you made me to recognise again that a single move can change the whole game , what i did was 2^x = 8^x - 4^x 2^x = 2^x ( 8 - 4 ) 2^x = 2^x+2 And what I got is 2=0 Then just want to make sure if my steps are correct , I got into video and then I noticed that you had written 2^2^x and 2^3^x as ( 2^x )^2 and (2^x)^3 , and just made it into an equation , and solved it , just a single step made wonder here , thank you so much next time if see this kind of problem I would try this method thank you

Thanks for the explanation, greetings from Colombia. A hug

You can also immediately do: x = log2(1/2+1/2sqrt(5))

@onlineMathsTV

Жыл бұрын

Nice but can you pls threw a little light on this procedure so that I can get the angle you are coming from?

@andersoncunha7079

11 ай бұрын

@@onlineMathsTV I think he used the fact that 2^x = n log2(n) = x

@Ideylaughintongues

10 ай бұрын

He simply used the change of base formula. We use that to condense logarithmic expression given in quotient form. In addition, no need to use a calculator to find the approximate solution. Keep it in compact form. That gives the EXACT solution.

@tontonbeber4555

9 ай бұрын

Yep, same for me ... took me a few seconds to solve mentally

The root corresponding to the non-principal value exists in complex space whereupon x = i.Pi + ln[1+sqrt(5)]/ln2 -1

私は日本人です 話している言葉は分からなくても数式は共通であり、数学の楽しさを共有できたことが嬉しいです

@onlineMathsTV

11 ай бұрын

Thanks for sharing in this joy sir. We are glad for this wonderful testimony sir. We all @onlinemathstv love you dearly❤️❤️💖💖💕💕🙏🙏

@HWR71

8 ай бұрын

The language is a kind of "English " 😅

@cmezion

8 ай бұрын

Да, математика это универсальный язык)

Logbase2(1+sqrt5) -1

👍👍👍

Sir, your videos are very interesting. Best regards from India.

Great....

nice 👍

This is great

I love this approach ....👍👍👍

@onlineMathsTV

Жыл бұрын

Wow!! We are glad you gained values from this video. Love you 💖💖💖

You're a extraordinary mathematics master i respect you. According to me who do know the math he could do the everything in this world....

@umeshhazarika4457

8 ай бұрын

Nice

Best tutor ever🙌

@onlineMathsTV

Жыл бұрын

Thanks a million ma.

@AmadaVictory-si5bc

Жыл бұрын

Sure, he is really trying.

I recall K-Sat problems ❤

❤❤❤phương trình mũ hay đó

Good , fun to follow explanation. Thank you.

Very nice video overall. 😊

@onlineMathsTV

10 ай бұрын

Thanks for watching

数学是观察👀的艺术，特别是回归到最基本的形式

Show!!!

Master Jakes on it again... Nice one sir.

@onlineMathsTV

Жыл бұрын

Smiles.....thanks

Same also just a suggestion

It's amazing how often the golden ratio appears.

I like to divide the original equation by 2^x this gives you 4^x - 2^x -1 =0, from there you substitute u=2^x => u^2-u-1=0 and apply the quadratic😁

@onlineMathsTV

Жыл бұрын

That is great also. Thanks 💕💕💕

@Yes_I_c4n

11 ай бұрын

Yep, just don't forget at the very end that 2^x = 0 could have a solution as well. In this particular case 2^× is always greater than 0, and so there's o solution to that, but genery you could have a solution. Once we get rid of something along the way, we tend to forget it has ever existed. 😊

@tontonbeber4555

9 ай бұрын

@@Yes_I_c4n Everybody knows 2^x=0 has no solution.

@tontonbeber4555

9 ай бұрын

Or directly finding u^3-u^2-u=0 and dividing by u as u=2^x is never 0

is there a faster method than this, this was like a whole page dude

Why did you use Ln? Log base 2 is more appropriate. Answer is X=Log(base2) of (1+sqrt(5)) - 1

Another great video from OnlineMathsTV. Pls kindly make some videos on WAEC past questions as the exam is fast approaching sir. Thanks sir.

@onlineMathsTV

Жыл бұрын

Your request is noted ok and we will soon make some videos on that, kindly give us some time. Thanks

Whether it is okay to solve this equation like this 2^x + 4^x = 8^x 4^x = 8^x - 2^x 4^x =( 2^x )^3 - 2^x 4^x = 2^x [ (2^x ) ^2 - 1 ] 4^x/2^x = 4^x - 1 2^2x -x = 4^x - 1 2^x - 4^x = -1 4^x - 2^x = 1 (2^x)^2 - 2^x = 1 2^x [ 2^x - 1 ] = 1 2^x - 1 = 1 / 2^x Log ( 2^x) - log ( 1 ) = log ( 1/2^x ) Log ( 2^x ) - 0 = log (1) - log ( 2^x ) Log ( 2^x ) + log (2^x ) = 0 2 log ( 2^x ) = 0 2x log ( 2 ) = 0 X = 0 Please correct me , if any of the step is wrong

I did this in my head in half a minute as soon as I realised it was 1/a + 1 = a, I knew that the solution is the golden ratio phi, and if 2^x = phi then x = Log (phi) to the base 2

@user-hl8mo6vi4q

8 ай бұрын

what is a?

@ggn4madhavmunjal985

7 ай бұрын

@@user-hl8mo6vi4q a is 2 to the power of x

You can also divide both side by 8^x

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❤❤❤❤

Обычное задание из первой части ЕГЭ.

Negative infinite

I short cut methods

Thanks

@onlineMathsTV

10 ай бұрын

You are most welcome sir and thanks to you for watching our content and leaving a comment behind sir. Much love ❤️❤️❤️💖💖💕💕💕💕

If you have a group of math please take me its link

Suggest: 4^x + 6^x = 9^x

ln(golden ratio)/ln(2). Formulaic.

Why we can't take 2^x = (1-√5)/2... Please tell 🙏

@xjoo92-fan

6 ай бұрын

真数条件より 2^x > 0 よって 0 以下の値である (1-√5)/2 をとらない

How did he get 2 to power of 3 from the 8

@babeluba007

8 ай бұрын

8=2*2*2=2^3

We can write the equation as 2^x+(2^x)²=(2^x)³ We can divide by (2^x) as 2^x≠0, in fact 2^x>0. Weget (2^x)²-(2^x)-1=0. Note that 2^x=golden ratio. Taking logarithm x=½×golden ratio

Is the answer zero?

Please correct the following. Ln (1+root 5/2) is not ln (1+root5) - ln 2

Her terefi 2^xbolsek

🙃

You can do it in 1 minute. 2 to the power 1 + 4 to the power of 1. Thus x=1. 8 = 2x4

@amadeolopez76

7 ай бұрын

2 + 4 does not equal 8

No entiendo bien el inglés. Creo que no has justificado por que solo hay una solución. Quizá lo has hecho y yo no me he enterado.

@onlineMathsTV

9 ай бұрын

There are other numerous solutions including the imaginaries ones but for the sake of this video tutorial we only solve for three. Thanks a million for watching and dropping a comment sir, much love from all of us here....❤️❤️💕💕😍

When u=0 value of x is not derived. Not even considered. Why?

@nicolomonti3655

8 ай бұрын

2^(x) is never equal to zero so it is impossible and he didn't consider it

Last step also can xlog 2=log 1+sqrt5/2

2^x + 2^2x = 2^3x t + t² = t ^3 Wrong: t1 = 0, t2 = -..... t3 = (...) 2^x = t3 2^x = log(2)t3 2^x = 2^ans X = ans(of the log)

You lost me at 5:52.

You going to far to find the value of X Make it simply as its value

@onlineMathsTV

11 ай бұрын

Noted.

2 x=-1

Immediately I find it to be just u**3-u**2-u=u(u**2-u-1)=0; just u can't be 0, plus u can't be negative. Is it really a Math Olympiad problem? Most high-school kids will solve it easily because nothing tricky or special...

X= ((ln(1+5^0,5))/2)/ln2

hold my beer...... x=0

Не желаю вникать в тарабарский язык

Por favor: éste x^3-y^3=xy+61

would be good to have it in English

2^x+ 4^x = 8^x; 〖(2*4)〗^x = 8^x; = 8^x; = = 8^x para cualquier valor entero de x > 0. Para qué tanto lio