This is why I'm going back to watching cooking videos.

At the end you can simply apply the definition of logarithm... 2^x = 5 ---> x is the power to wich the base 2 must be raised in order to obtain 5 so you can write: x = log2(5)

I feel the quick way to approach this problem is to recognise that we have a polynomial where both y^3 and y exist. This should ring a bell that we should find some cubed number somewhere. We can write 130 = 125 + 5 = 5^3 + 5. That's it. Immediately we get y^3 - 5^3 + y - 5 = 0, or (y -5)(y^2 + 5y + 26) = 0, where y = 2^x. Here onwards, it is simple.

130=26×5=(5^2+1)×5=5^3+5; 8^x+2^x=(2^x)^3+(2^x); 2^x=5 My high school math teacher used to tell me,to understand an equation better you need to dis-simplify the brief side other than to simplify the complex side.

Me given IIT 20 yrs back and working in IT from 15+ yrs still willing to learn this so that I can teach my kid. Feels like back to square one. 😂 happy learning guys.

Instead of using decimals you could have used more logarithmic identities when you were checking. I think that would have been cleaner.

Change of bases can be used to simplify the final expression 8^log_2(5) = (2^log_2(5))^3 by associativity and commutativity =5^3 since x^log_x(a) = a =125 And the same with the other term, so 125+5 = 130

We will also have complex values of x... Since we have a cubic polynomial y^3+y-130=0, it will therefore have 3 roots out of which 1 is real (log5/log2) and other two are complex Other two values of x hence will be: x=log(5+- isqrt71)/2/log2

I loved my Math's Teacher Sir A Hameed Wayn, in Metric Class ... And after him, you are the 2nd one, whom I would like to praise .. Since my First Teacher changed the School , and just for Mathematics I followed him to that school. From there, you can see my love for Mathematics . Liked your V-Log ... Though i dont know your name .

Just subtract 130 from both sides and solve for the x-intercepts of the equation 8^x + 2^x - 130 = y. When y= 0 the graph intercepts the x axis.

For all functions of type f(x)=ax^3+bx+c the equation f(x)=0 can have only one real root in case of a>0, b>0, because f'(x)=3ax^2+b>0 and therefore f(x) is monotonically increasing. Also, if the equation f(x)=0 has rational roots in the form p/q, p is a divisor of c and q is a divisor of a. If a=1, the rational roots are always whole numbers. One can immediately see that in the specific example f(y)=y^3+y-130=0 , y=5 is a solution, because we first check the whole divisors of 130 (+-1, +-2, +-5, +-10, +-13). Try to solve x^7+x^5+x^3+x=170 with your method....No chance. With above, you show that x=2 is the only root in just 2 rows.

I remember fondly a time where maybe, just maybe, I might have had some idea, but I haven’t used anything beyond grade 9 algebra in 20 years.

All these problems seem a little too easy for olympiads

I do not yet have the mathematical experience to have come up with that line of thinking to jump to thinking of the factorisation of 130 then rewriting as factor by grouping. I got to the step where I substituted u = 2^x but had no idea that was actually the way to proceed. I was stuck at what to do now with the 130 as I had no obvious way to factorise u^3 + u - 130 = 0.

So hard. Congrats. Several math concepts 👏👏👏👏👏

I did it like this... 2^(3x)+2^x=130 Let 2^x=y Then y³+y=130 By observation, y=5 Hence 2^x=5 X=log(2)5, ie log 5 base 2 Overall, this problem was on the easier side if it's from olympiad, as I and my friends (Indian high-school students) were able to solve it pretty easily 😅

Thanks you so much for refreshing my memory from 10 years ago, Now I wish I have continued on the math field instead

Ótima explicação. Mas, quando você já havia encontrado que 2 elevado a x era igual a 5 , já poderia ter usado a definição de logaritmos e chegar direto na conclusão que x é igual a log de 5 na base 2 . Ou fazer assim seria um erro matemático? Parabéns pelo excelente vídeo!

Having got to y³+ y = 130 it's not hard to try a few small numbers and see that y=5 is a solution. From that the quadratic part could be worked out, though it should be obvious that there can be no other real solutions since y>5 would be too big and y

Let a=2^x, then a^3+a=130, so a=5 is a solution, then x= ln5/ln2. The other solutions can be found by factoring out (a-5) and solving the quadratic.

This is why I know anatomy so well!

Confuse the number 26y - 25y, can you explaine it!

4:50 Redundant. By definition of logarithm, it is the value of the exponent to put on the base to get the argument of the logarithm. So basically in this case x is essentially, in base 2, log(5).

Incredible how much effort you put in this. 130=5³+5¹ and this is identical to your y³+y. Therefore y = 2^x = 5. => x = log2 (5). Easy.

FASTER APPROACH : 8^x+2^x is always increasing function, hence one root only. Setting x equal to some numbers, we realize that x is between 2 and 3. Consequently,y is between 4 and 8, but y is obviously less than 6 by analyzing cubic equation. So y is between 4 and 6, try 5 and we get that y=5. Everything can be solved in mind.

Спасибо! Не понимаю концовку с таймкода 4:38, и так ясно, что логорифм - это степень числа по основанию.🧐

In solution checking a^(log c to base a) can be written as c^(log a to base a) . So 8^(log 5 to base 2) can be written as 5^(log 8 to bas 2), which is 5^3 and 2^(log 5 to base 2), is 5^(log 2 to base 2) which is 5 5^3+5=130

First observe that 8^x=(2^3)^x=(2^x)^3. Then set 2^x=y and you get the equation: y^3+y=y*(y^2+1)=130. A few trials gives y=5 and thus x=ln(5)/ln(2)=log(5)/log(2).

I think the question is ill phrased because the types of the symbols in the equation are not specified. For example: no solution over integers. One solution over reals. 3 overvcomplex numbers. But what if we consider prime fields F_p with p>130 or any other algebra type where those symbols could be interpreted in?

Was in it better to approximate x between 2 and 3 at a glance?

It's too complicated, I mean after getting 2^x = 5 we get in accordance with the definition of a logarithm x = log2(5) where log2 is a logrithm with base 2. All following calculations in the video are unnecessary.

Your handwriting and logical thinking ability are awesome ❤

reminder me my school years in one of the best math schools in Russia 25 years ago, now all forgotten but still these problems are solvable on the fly almost🙂 good times it was

very informative and easy way of teaching

To complicate solution We can divide by 2^x and y= 2^×

How does this relate to real world practicality, thank you.

Really crazy how easy this problem would be for my past self studying engineering. Being done with school and doing the same shit over and over again at work really rots your brain

Your voice is so soothing

If you are asked for only the Real solution then youbare correct. If you want all solutions, you must take the quadratic into account and get two more Complex solutions.

For 2^x=y y³+y=130 y(y²+1)=130 Clearly for y=5, equation satisfied so x=log5(base 2) is the correct answer.

Anyone know which pen shes using? Looks so smooth 🥹

The longest way to do that equation.

Great session

Aah my favourite maths algebra those days ❤🎉

how can we split the middle term like this . u wrote 26 and 5 then how can you take 26 and 25 in tge next step

y(y²+1) = 5(5²+1) ...at this stage it should be apparent that y = 5 🤪 Given y = 2^x Then 2^x = 5 log²(2^x) = log²(5) x = log²(5)

Yes 130=26x5, but how to get y=26y-25y ? I mean how to think of such an inference ?

Instead of that lengthy solution, let 2^x=y, and then let f(y)=y^3+y-130 For y=5, f(5)=0 f'(y)=3y^2+1, which is positive for all values of y, meaning f(y) is a monotonically increasing function, which makes y=5 the only root of f(y) Then 2^x=5 and solve using logarithmic properties

Why aren't the imaginary solutions taken into account? I mean, they are solutions to the original equation, aren't they? Is there an assumption that we must find the real solutions only?

I think you could simply write 2^x = 5 => x = log2(5) without all these log divisions, because log2(5) literally means power to which we have to rase 2 in order to get 5, which is x in our case

To all the pretentious people who keep commenting that she made it hard, please note that she is trying to teach everyone at any level, that's why she chose the most straightforward method that any type of student can follow. She is not teaching only the undiscovered geniuses such as yourselves.

Nicely done. Weird how your 2's are written differently even in the same equation

This can be solved by vanishing factor method By putting y=5 y-5 is a factor 5^3+5-130=0 y^2(y-5)+5y(y-5)+26(y-5)=0 (y-5)(y^2+5y+26)=0 y-5=0 or y^2+5y+26=0 y=5. D for quadratic equation 5^2-4×1×26 is less than 0 No real roots. Solution will be y=5

Use substitution. Substitute 2^x =y and solve polynomial equation.

What happened in the end? Why the author used approximately solve? 2^log(2 5) = 5 by log's definition. And 8^log(2 5) = 2^3log(2 5) = (2^log(2 5))^3 = 5^3 = 125

i solved it like this after 2^x(2^2x+1) = 130 so taking factor of 130 as two terms like 65X2 or 26X5 or 13X10 and trynna see which of these two terms satisfy that eq which is on taking 2^x= 5 so thats the answer after taking log of this

은가은씨 오늘도 즐거운 불금 되세요.....

You make the solution to obtain the real part seem too complicated. from the point where (2^x)=5 by just using the conversion law from Exponents to log we obtain Log5_2 = x. Even to check the result, using logs instead the decimal produces a more exact result or (125+5=130).

2^log²5 = 5 (by definition) 8^log²5 = (2^3)^log²5= (2^log²5)^3= 5^3=125 There is no need to do approximete calculations.

Use rational zero theorem. Y=5. And divide the function by (y-5) and get the quadratic. Much faster....

Beautiful!

Waiting for more videos, Mam

I have simplier solution. Multiply both by root x, so then 8+2 = x base root of 130. Which easily = 2.3....

Yeah I got it too. But when you check it, you should not convert to log value and you can simply find it.

Do you have a short cut to solve this....back to 1994 my tutor can sovle this in 2 minutes

4:47 x is already log 5 to the base 2 as per logarithm definition,. Isnt it

Why cant you just add the left side to 10^x and do logarithm? Did I forgot of the basics of math? xD Please answere me I am really curious.

I got from a much easier and faster way 2*3x and 2*x should be equal to 130 so the sum of two of 2 power something should be equal to 130 which is close to 2*7 hence if x=2 the answer is 64+4 and if x= 3 it would be 528, therefore, x should be between 2 and 3 and closer to 2 than 3 and 3x kinda feels be close to 7 as 128 is close to 130 so 7/3=2.33 and as it wouldn't add up (cuz 127+4,9 is greater than 130 we should decrease 2,33 by 0,01 per time to find the right answer then we got the answer of 2.32

I'm already confused at 1:29 . Where did the 25 come from?

Although I knew the answer to the problem, being lazy I preferred "Hit and Trial Method" from the given option for these type of questions. This saves a lot of time 😅😅

Thank you 🥰🎉👏

Let y=2^x then y³+y-130=0 Easy to see y=5 is one solution. I will skip complex solutions which just requires synthetic division and quadratic formula to solve. 5=2^x -> log_2(5)= x -> x= ln5/ln2

I looked at this for like 2 minutes without a thought of any complex maths and thought the answer might be X= 2.25 and I’m honestly pretty pleased. Lol

I love these, I do data analysis for word and they have no practical application but I still love them

If we simply check in starting of cubic equation that one of its root is 5 as 5^3+5-130 equals zero, it can me much easliy solved

Wawoooo. So simply explained

Can someone show how to solve it with a graph?

How can 130 be 26x 25

This is hard core algebra. Cubic equations, quadratic equations w/ the quadratic formula, logarithms, fractional exponents, ect…. I did stuff close to this level in high school. The difference was that it did not have multiple layers of this complexity. My takeaway from that experience was that algebra wasn’t hard so long as you worked a lot of different problems and got plenty of practice.

I didn't get the last thing you did aa y-5= 0. If we can do this, then can't we do this for (a+2)(a-3)= 14 as a+2= 14 and a-3= 14? And that would give a completely wrong answer as the value of a should be 5. Please explain this.

How 25 came during factorisation?

I have some concerns.......

I learn my level is olimpic thanks to this channel.😂

Before watching: This is not one that can be easily solved by simply plugging in integers and hoping for a result. Our solution is going to be somewhere between 2 (8^2 + 2^2 = 64+4=68) and 3 (8^3 + 2^3 = 512+8 = 520), and probably closer to 2 than to 3. Therefore, we have to actually do some calculations. Alright, 8 = 2^3, and (a^b)^c = a^(bc). Thus 8^x = (2^3)^x = 2^(3x). We declare U = 2^x. Then 8^x = 2^(3x) = u^3. Then we have u^3 + u = 130. Subtract 130 from both sides to get u^3+u-130=0. Now, we attempt to factor this. The factors of 130 are 1, 2, 5, 10, and 13. Of those, U=10 gives us results far too large, and U=2 gives us ones far too small. U=5 gives us 125+5-130 = 0, which is accurate. Thus, (u-5) is a factor. after doing some division, we can factor the equation into (u-5)(u^2 + 5u + 26 )=0. Going to use the quadratic formula on the second factor, we note that the discriminant is negative. Thus, these are not real roots, so we can skip this section. Thus, we will go with U=5. However, we're not done! That's the solution for U, not for X. U = 2^x. Thus, we have 2^x = 5. Take log_2 of both sides: Log_2(2^x) = log_2(5) -> X = log_2(5) Log_2 of 5 will be between log_2 of 4 (2) and log_2 of 8 (3), likely closer to 2 than 3. This checks out. If you want to change this to a different log using change of base, you may do so. log_b(a) = (log_c(a))/(log_c(b)),. Then using natural log ln, with a =5 and b = 2: x = (ln 5)/(ln2). (We're not doing this for the sake of precision, but rather so it can be easily checked on a calculator. A lot of calculators don't have functions built into them for logs of different bases, at least not ones that you can get to easily. Thus, you have to switch to either common log (log_10) or natural log (log_e))

I think 5 is a rather easy to find "obvious" solution by searching a integer which cube is close to 130. Then it becomes a polynom division. The proposed factorisation is nevertheless very smart !

Can't we just use log on both sides?

great video but the verification can be a little bit better if it's like this: 8^log5 base 2 + 2^log5 base 2 = (2^3) ^log5 base 2 + 2^log5 base 2 = (2^log5 base 2) ^3 + 2^log5 base 2 as we already know a ^logN base a = N => (2^log5 base 2) ^3 + 2^log5 base 2 = 5 ^3 + 5 = 125 + 5 = 130 excellent video keep it up and upload more videos like this.

I could have just put some value in place and solved mathematically to make a guess that i have to go up or down a little. Than three attempts later i might have got x=2.3 I have done it many times. You dont need to go in multiple algebrac complications to solve a problem this simple. Afterall numbers are used to avoid hit and trial of actual objects. We can use numbers for hit and trial as that wont even use extra ink or time cz the equations are more complex than hit and trial. Smart work is also intelligence!

Please I don't understand why you multiplied the 5y by 5 to insert into the equation

Go easier by using Factorizing with polynomial and basic logarithm

As soon as she got it to Y Cubed plus Y = 130, the answer jumped out at me and I yelled it out loud, trying to beat her to the punch. Who knew I actuality had time to go check the mail, first?

Nice, but you should see, that y^2+5y+26=0 can't have any real solutions without computing by seeing, that it's graph the shifted y^2-graph into positive direction.

I think the easiest solution is that we know 2^7=128 next is 8^(1/3) = third root 8 =2 so 128+2=130

1:37 is confusing? Why and how can you write y=26y-5y?

We can do it faster. 130 has only 3 non repeating prime factors 5, 13 and 2. So there is not much of a substitution job to check which one satisfies Y * (Y^2 +1) = 130 And its 5.

Beautiful question. It's answer is log5/log2 both on the base 10 👌👌

Love this, Thanks ❤

What is this, olympiad for class 5?

In real exams, you can just substitute the given choices if satisfies the 130. It will save lot of time instead of solving.

Log 2^x = x log 2. Please learn this. For your solution: X= log5/log2.