Solving a Non-standard Exponential Equation

This video is about solving an Exponential Equation
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[@yourmathtutorvids]

Great video! You always pick questions that look simple but require a little bit of thinking 😂

[@SyberMath]

Good catch!

[@aashsyed1277]

0 LIKES>?

[@mega_mango]

@@aashsyed1277 99*

[@snejpu2508]

Nice problem. Fun fact: sqr(3)/2 and 1/2 are well known values for cos(30) and sin(30). If you square those values you get 1.

[@GenZClub]

Wow great approach

[@SyberMath]

Yes, thank you!

[@DaveJ6515]

That's the way to see it at once. That and the fact that when the base is in ]0,1[ the power is monotonically decreasing, the sum of monotonically decreasing functions is still m.d., hence solution is unique.

[@sueyibaslanli3519]

Yeah , that is what I think

[@siddhartham1973]

@@SyberMath u say u teach maths n u r agreeing wuth that is that how clear ur basics r mister

[@pranobg]

Put x/2 = t. You get 4^t = 3^t + 1, or 4^t - 3^t = 1. As LHS is a strictly increasing function, the only solution is t = 1 (or x = 2)

[@frentz7]

All good (and nice and elegant!) except the part where you guess the answer ..

[@sundeep0207]

My method using basic calculus. Given equation is 2^x = 3^(x/2) + 1. If x log2 (2^x - ((root 3)^x)) As x>0, (2^x - ((root 3)^x)) > 0. Therefore, f'(x) > 0. This is strictly monotonic function in (0, ♾️). Hence, if a solution exists, it will be unique. From observation, we can see that x=2 works. And that's the only solution. Edit : I tried solving on my own before looking at the video and posted mine here. This is the pretty similar to what's in the video at the end. Apologies if you find reading my solution to be a waste of time.

[@SyberMath]

I like it!

[@mchowdhury7708]

Nice! My solution is quite similar to yours but I did it a bit more intuitively. As you mentioned x

[@MelomaniacEarth]

Great work📌this should be pinned 😁✌

[@kerbicz]

Another solution using elementary calculus. Substitute _a:=x/2._ Then our equation reads as: _4^a-3^a=1._ By the Lagrange formula applied to the function _t-->t^a_ we get _4^a-3^a = ac^(a-1)_ for some _c_ between _3_ and _4._ But the right-hand side of the latter equation is greater than _1_ when _a>1_ and smaller than _1_ if _a

[@sundeep0207]

@@kerbicz I am not familiar with this lagrange formula. Seems interesting. Can you please point to some resource to learn.

[@thomaswaynejunior3110]

2ˣ - 3^(x/2) = 1 = 4 - 3 = 2² - 3¹ 2ˣ = 2² x = 2 3^(x/2) = 3¹ x/2 = 1 x = 1 × 2 = 2 since both have the same result, thus x = 2

[@fedorlozben6344]

2^x=3^(x/2)+1 2^(2y)-3^y=1 4^y-3^y=1 The value for y is only 1,else lhs grows too fast or too slow That is why x is 2...as so easy!

[@SyberMath]

Nice!

[@leickrobinson5186]

That’s what I noticed, too! :-D

[@victorchoripapa2232]

Very nice solution

[@mohamadabbas8384]

Sehr schön

[@jin_hyukkiee7469]

I looked at it. And just said 2 😂

[@proverbalizer]

Exactly what I did. Guess and check. 2? Yep.

[@gamergamer1296]

This is actually a good method to solve it, you put everything in one side, you name it as a function and then "guess" The solution and if the function is 1-1 you basically prove that this is the only answer

[@sahilsinghbhandari444]

As we can clearly seen that there will be one solution for x=2 by substitution, but the checking method for x>2 and x

[@SyberMath]

Thank you!

[@alexbarac]

A very nice way to find the solution x=2! What I did was simply observe that 2 is a valid solution, and then prove that f(x) = 2^x-3^(x/2)-1 is an increasing function, which leads to the same conclusion that the equation only has one solution which is 2.

[@ryzenandrewgarcia657]

Great video! I tried it only by inspection knowing what 3 and 1 would sum up to a square number, so I took the chance to regard x as 2. 😅

[@ndmitry534]

we analyze the function, make sure of the uniqueness of the solution, the extremum point, determine the sign of the derivative, limit the range of solution zones and select x = 2

[@SyberMath]

Sounds good

[@ndmitry534]

@@SyberMath This is how I was taught. If you don't find a way to a solution right away, use the mathematical apparatus, analysis of functions. It always helps.

[@yusufjamal5474]

1 in the equation can be written as (2+sqrt(3))(2-sqrt(3))=2^2-(sqrt(3))^2.On taking 3^(x/2) term on the left side ,and comparing the powers will give x=2.

[@duraace6511]

I used a different approach: cos^x(pi/6) + sin^x (pi/6) = 1 if and only if x=2.

[@SyberMath]

Nice!

[@addaboualem3222]

thank you for your explanations and your better understanding

[@SyberMath]

You are welcome! Thanks for watching!

[@oppayongvlogs1285]

This is my Major Subject. Bsmathematics from the Philippines. Great explanation

[@SyberMath]

Glad you liked it

[@davidgillies620]

If you set f(z) = (sqrt(3)/2)^z + (1/2)^z with z complex, the level set of |f| = 1 in the complex plane is a rather nice periodic curve.

[@SyberMath]

Nice. What can we use to graph it?

[@feniceazzurra7168]

@@SyberMath graph

[@marcushendriksen8415]

@@feniceazzurra7168 yeah, graph. As in representing the function as a set of ordered pairs on a plane

[@AlephThree]

I wrote LHS as 4^(x/2). We have 4^(x/2)>3^(x/2). Can then show expression works with x=2 and that LHS>RHS for x>2. Just realised I may have missed a root for x

[@HemantPandey123]

4^x/2-3^x/2=1. Put x/2 = y. Means 4^y-3^y = 1. Implies y=1.Expand using p^n-q^n= (p-q)(p^n-1.q+...q^n-1.p), y=1 is only solution. Further we can use graph to prove 4^y and 3^y+1 meets only at y=1. Also you can take sqrt(3)/2 = sina and (1/2) = cosa and use sin^2a+cos^a = 1 to prove x=2 is only solution.

[@SyberMath]

Good!

[@septembrinol1]

The idea of using trig substitution here is very nice

[@priyankgupta]

Love your channel. Severely underrated. Please keep at it, you will grow big one day.

[@SyberMath]

Thanks, will do! 💖

[@randyklijn6267]

Nice explanation! Really like the feeling of understanding those complexer math problems :)

[@ritampaul5005]

Very good problem. Awesome explanation specially the graphical approach.

[@SyberMath]

Glad it was helpful!

[@shreyan1362]

I see your channel gaining 1k views on this video alone everyday, that's nice

[@SyberMath]

Thank you!!! 🥰

[@michelkhoury1470]

Very nice solution I've got also x=2 as a solution to the equation since 1 = (3/4 + 1/4)^(x/2) = (3/4)^(x/2) + (1/4)^(x/2) And we know that (a + b)^n = a^n + b^n is true for only one value : n=1 Therefore the equation is only true for x/2 = 1 i.e. x=2

[@SyberMath]

Thank you

[@nonoobott8602]

Nice problem. Amazing solution. Thanks for sharing

[@hcadambi1]

If you recognize sine and cosine function for the values that you see, then x=2 is the solution knowing (cos60)^2 + (sin60)^2. =1

[@SyberMath]

Nice!

[@hcadambi1]

Thanks 🙏

[@phatmanoch3699]

sec(30) = 2 tan(30) = 3^(1/2) equation can be written as (sec30)^x = (tan30)^x +1 trigonometry identity: (sec A)^2 = (tan A)^2 +1 we get x = 2

[@pedromendes6846]

There's a much easier way. 2 = 4^(1/2), therefore 2^x = 4^(x/2), then that equals to 3^(x/2) + 1. If it wasnt for the x/2, you'd have 4 = 3 + 1, so the only way to keep the base number the same is by making x/2 = 1, and we get x = 2

[@SyberMath]

Interesting

[@alirezamirzamani9009]

This shows that there are limitations for the use of exponential functions The way that sometimes even Logarithmic ways can not work out .

[@frentz7]

Yes! .. great comment. It is not clear how to solve this kind of problem.

[@WolfgangKais2]

„Getting reminded“ of something nice isn‘t more systematic than „guess and check“.

[@SyberMath]

True

[@abhinavdiddigam2330]

We can give a reason to why x=2 is solution √3/2 is sin60 and 1/2 is. cos 60 and hence sin powerx 60 + cos powerx = 1 which reminds us of the infamous identity Sin^2 theta+ cos^2theta = 1 And hence X=2

[@SyberMath]

That's right

[@seegeeaye]

To prove the solution is unique: 2^x - (sqrt 3)^x is a monotonically increasing function which intersects y = 1 at only one point

[@chorocknamu8481]

I'm 14years old from Korea.To he honest,in Korea most peole who are about my age can solve this in 10seconds.Korea is VERYVERY EDUCATIONAL nation.

[@한지훈-q8n]

divide as 2^x 1=(sqrt(3)/2)^x+(1/2)^x --> 1=sin(pi/3)^x+cos(pi/3)^x x=2

[@pranavmeshram3323]

If we reduce the eqn mod 3,we get that X should be even,i.e,x=2n...putting that in original equation we get,and bringing 1 to the left,by difference of squares we get, (2^n-1)(2^n+1)=3^n Only n=1 satisfies the above as all values of n greater than 2 always give a difference of more than 2 between the factors...n=1 implies x=2

[@SyberMath]

That's neat!

[@pranavmeshram3323]

@@SyberMath Thank you!

[@SyberMath]

@@pranavmeshram3323 You're welcome

[@jomarviesca2150]

In the video, 1/2 to the power of x is equal to the quantity (1/2) raised to x. What is your justification on this ? What if the numerator is not 1 for example ? Would it still hold true ? I'm quite confused. Thanks for your help.

[@TheMartian11]

I just used the identity that cos^x (a) + sin^x (a) = 1 only when x = 2. In this case a=30

[@SyberMath]

Very nice!

[@georgenguyen9722]

great

[@tonyhaddad1394]

Cooll approach !!!!! Good job

[@SyberMath]

Thank you!

[@anhchuvuive8146]

When a^x ( a > 0 ) then put the x down to the equation . We have x + 2 If a^(x/b) then I make this formula : a^(x/b) = ( b^roots(a))^b*x - (b^roots(a))^b ( when a > 0 ) If a < 0, then a^(x/b) = ( b^roots(a))^b*x + (b^roots(a))^b Now we can solve this equation : 2^x = 3^(x/2) + 1 x + 2 = (sqrt(3))^2*x - (sqrt(3))^2 + 1 x + 2 = 3x - 2 -2x = -4 x = 2 ( the answer is x = 2 )

[@-basicmaths862]

2^x=3^x/2 +1 2^2t=3^t+1 4^t=3^t+1 4^t-3^t=1 Difference is one so 4^t and 3^t are adjesant integers. t=1 x=2

[@SyberMath]

Nice!

[@autumnstar7346]

I only saw the thumbnail nail and decided to solve before even watching your video. My high school math teacher must have taught me real good. Math isn’t even my string point, yet I just looked at the equation, and in just a couple of steps with a lot less writing, using just basic knowledge of solving for x, and keeping in mind the rules for negative exponents and fractions I solved for x,and got x=2.

[@technowey]

I looked at it, asked what powers of three are one less than a power of two, and immediately saw the answer. Estimated solution time, 2 to 3 seconds. Then I wondered if there is more than one solution.

[@SyberMath]

Nice!

[@lambertquesada1363]

Sybermath did that also I believe but for the sake of the viewers, he needs to explain why the answer is only 2...

[@frentz7]

Good job! Now let's use what we learned from the video: try 2^x = 5^(x/2) + 1

[@iwd1705]

I would have started with 2^x - 1 = 3^(x/2). This would mean for x = 0 the lefthand side would be zero as well which is a great number, because the righthand side can never be zero or even negative. So for any x < 0 we're done. Then I would have substituted t = x/2 which would have given me 2^(2t) - 1 = 3^t. That expression can also be written as 4^t - 1 = 3^t. Easy to see there is a solution for t = 1 which means x = 2. And it has to be the only one because standard exponential functions as 4^t or 3^t are standardly increasing from left to right and 4^t will increase more than 3^t when t>0 but less when t

[@aashsyed1277]

the best channel! thanks so much!!!!

[@SyberMath]

Glad you enjoy it!

[@GerardWassink]

I got x=2 after looking at the ‘riddle’ for five seconds…

[@enejidjsi5939]

But how do you know if there are any more solutions.

[@GerardWassink]

@@enejidjsi5939 That was not the question. The question was “are you able to solve this”. I did.

[@blessos]

@@enejidjsi5939 it's clear by plotting the curves for the two exponentials

[@mrinalkantibose3504]

Very nice approach, amazing.

[@SyberMath]

Many thanks!

[@thomaskim5394]

The function has to be continuous and decreasing. Then, the intermediate value theorem....

[@Asycoworld]

i subscribe Ur Chanel bcs i solve this problem in different way but solutions were same ..and i think i can get more best problem and equation from this channel.

[@SyberMath]

Welcome aboard! 💖

[@RifatRahmanRimon]

This is how I solved it. I let, x=2p. 2^(2p)=3^p+1 4^p-3^p=1 Now, let f(p)=4^p-3^p If we see, f(-p) is not an even function, neither f(p) itself is a periodic function. Hence, any value other than p=1 will not result 1. So, x=2p=2

[@riazwazir1592]

I think Taking log on both sides and will be solved directly.

[@umkamaks3296]

Its was easy, answer 2! I counred in my mine

[@manuelgonzales2570]

Brilliant!

[@SyberMath]

Thank you! 😍

[@TheJJ65]

another way of showing that 2 is the only solution is that sqrt(3)/2 and 1/2 are cos and sin of pi/6 respectively, and considering the sin^2(x)+cos^2(x)=1 identity, you can figure out that the answer is two and it's the only exponent where that identity works.

[@SyberMath]

Very good!

[@Shreyas_Jaiswal]

1:00 Seeing here came to my mind that x can be 2. And that's the only answer.

[@woowoo8918]

That's great solution!, I understand the equation and your voice are good!!! 2^x=3^(x/2)+1 2^x=sqrt(3)^x+1 2^x/2^x=(sqrt(3)^x/2^x)+(1/2^x) 1=(sqrt(3)^x/2^x)+(1/2^x) (sqrt(3)^2)^x+(1/2)^x=(2/2) x = 2, (3/4)+(1/4)=4/4 = 1

[@SyberMath]

Thank you!

[@זאבגלברד]

The function y=2^x is a "basic" function. Also (3^0.5)^x. If we solve where they intersect we get x=0. According to the drawing it is clear that the obvious solution x=2 is the only solution.

[@SyberMath]

What do you mean by basic?

[@זאבגלברד]

@@SyberMath Functions that a high school student needs to know their drawing, by heart.

[@dangvuthai]

So, for any problem of this format a^f(x) + b^g(x) = c, if it's solvable by hand, it should be able to be converted to (sinm)^h(x) + (cosm)^h(x) = 1. Does that sound right?

[@frentz7]

Very well articulated! but can we do it?! .. the numbers in this equation were designed to make this possible. for example, try 2^x + 3^x = 1. somehow I don't think you will manage to stuff this into the "sin m" and "cos m" format.

[@mackenziekelly1148]

Just say that 2^x=4^(x/2). Thus you get 4^(x/2)=3^(x/2)+1. It's evident (since 4 will outgrow 3 exponentially) that this is only true when the exponent is 1. Therefore x=2. Done.

[@SyberMath]

Very good!

[@7pimath199]

Interesting video, nice solution liked it, particularly the part that you prove that there is only one solution. Actually, I'm used to preferring the change of variables technique and solved this one by using a change of variables technique twice. You can refer to the video on my channel. (x/2)=a, then using logarithms and then 3^a+1=b, then find b, then a, and then x :D

[@SyberMath]

Thank you!

[@frentz7]

How to solve 3^a + 1 = b ? I didn't understand

[@7pimath199]

@@frentz7 Hi, you can check a shorts video on my channel named Solving Exponential Equation - kzbin.infoLY-FowrC9lc?feature=share . It involves some repetitive steps of using logarithms, hope it will be useful)

[@johnmencias7743]

So my approach was slightly different ... Since 2 = 4 exp 1/2, then we can express 2 exp x as 4 exp x/2 Obviously, we can also express 1 as 1 exp x/2 Therefore, the equation becomes: 4 exp x/2 = 3 exp x/2 + 1 exp x/2 Let x/2 = y, then the equation becomes: 4 exp y = 3 exp y + 1 exp y But 4 = 3 + 1, so substituting in LHS: (3 + 1) exp y = 3 exp y + 1 exp y Clearly y = 1 is a solution, but since y = x/2, then x = 2 is a solution. If y is gt 1, then LHS is gt RHS If y is lt 1, then LHS is lt RHS

[@mr.no1120]

Here is the shortcut one: If you put 0 as a value of x then both sides are not equal...and same when you put 1 as a value of x But when you put 2 instead of x then both sides are equal in the equation so ...it means that 2 is correct 2²=3 raise to power 2/2 + 1 (2^x=3^x/2 + 1) 4=3¹ +1 4=3+1 4=4 So 2 is correct 2/2=1....by putting x=2

[@superluigi6423]

Me when the video started : the answer is 2 Me at the end : why he did all of this when the answer was easy 2 found ???

[@shreyan1362]

you always have to prove that there is no other answer 🙂

[@superluigi6423]

@@shreyan1362 yeah I know :)

[@SyberMath]

Well, finding one solution is always a good thing

[@tigranchilingaryan2599]

Thank You for video

[@SyberMath]

You are welcome

[@mdsahabuddin254]

Its really interesting 👍

[@angrytomato3365]

I think x

[@jimcharamis6903]

I would like to propose another solution. Firstly, one can easily observe that x=2 is an obvious root. We will attempt to prove that x=2 is the only possible root. The second observation is that for non-even x this equation cannot hold., 3^(x/2) or (sqrt3)^x can be an integer only if x is even. However, 2^x -1 is always an integer. Thus, it must hold true that x=2k. With this substitution, the equation can be written as 4^k =3^k +1. This is a standard exponential equation which obviously has only one solution, k=1 and thus x=2.

[@SyberMath]

Nice!

[@jimcharamis6903]

@@SyberMath Thanks :) !

[@muse0622]

y=a^x(0

[@SyberMath]

Very good!

[@muse0622]

@@SyberMath Thanks!

[@siralanturing9103]

The million dollar question is:- Does SyberMath's face resemble a bell- shaped curve or a curvy bell?

[@SyberMath]

😂

[@acrylista2130]

Your solution is great

[@SyberMath]

Thanks! I appreciate that

[@algalgod159]

Another way to solving it: start with x=0, see that it is Not a solution, then x=1 (same), then x=2 IS a solution, then because we have two exponential curves, they can only intersect in one point, hence x=2 is the only solution. correct me if im wrong

[@ahmedayad6338]

Hi .. Thanks for this beautiful video... If I may ask... what's the name of the software you used in the video.

[@SyberMath]

Sure. Thank you! It's Notability

[@georgeradulescu9443]

amazing stuff

[@SyberMath]

Thank you! 💖

[@fred8780]

2^x - 3^ 0.5x = 1 sandwich the answer between too high and too low. if x < 0 then y is negative; too low if x =0 then y =0; too low if x > 0 then y increases exponentially. there will only be one real answer. x=1 is too low; x = 3 is too high. at x=2; the denominator cancels out. at x = 2; 4-3 = 1;

[@mcwulf25]

Put X=2y then 2^2y = 3^y + 1 3^y = (2^y + 1)(2^y - 1) The two factors differ by 2 so only one can be a power of 3 while the other must be 1. Clearly the larger is a power of 3. 2^y - 1 = 1 2^y = 2 y = 1 So x = 2. That's it.

[@eeawah8649]

all these people be typing out solutions and my lazy ass didnt even wanna try

[@SyberMath]

😂

[@mouloudadjroud716]

لا حدث و لا شيئ، الغريق يسخر من اللذي يمسك بقارب النجاة...

[@tgx3529]

Nice idea

[@davutsauze8319]

Could you get other answers if you allow for imaginary numbers?

[@proverbalizer]

I just looked at it for 2 seconds, thought about something simple I could easily guess and check and picked 2. Done, lol

[@proverbalizer]

2^2=3^1+1 bingo

[@godisone6518]

Good

[@archilarkania7203]

It's trigonometry. You only get value like that in sin and cos.

[@jalajam1]

Sin^2+cos^2=1,sin^2(60,30)+cos^2(60,30)=1

[@iamrandomcrbot283]

I dont know why but that eazy solution made my brain at least 1.5 times bigger

[@jamesdunk9118]

This problem is much easier to solve: 1

[@SyberMath]

That's good

[@amehachewa4757]

you have to say to the public for such questions as extrenous answer unless the students can be confused because you did not get the answer mathemathically.

[@RexxSchneider]

Messy numbers. Use the substitution u=x/2 to clean it up. We get: 2^(2u) = 3^(u) + 1 4^u = 3^u + 1 v = 4^u - 3^u = 1 A lot easier to see solutions with integer bases and u=1 should be obvious. That gives x=2 immediately. The same logic as in the video applies to showing it is the only solution, although I think comparing powers of 4 and 3 is clearer. Since both 4^u and 3^u are exponential functions, we know that 4^u increases faster than 3^u when u>0, so when u>1, v must be >1. Similarly, as u goes more negative, we know that both (1/4)^(-u) and (1/3)^(-u) tend to zero monotonically, and their difference can never reach 1 (v tends to 0).

[@Galaka100]

Responding to Rex Schneider: Actually an easier next step is to substitute T=3^u so v=4/3T-T=1. Simplifies to T/3=1 or T=3. Therefore u=1. Therefore x=2.

[@RexxSchneider]

@@Galaka100 Sorry, but I'm not sure that's correct, because if 3^u = T, then it doesn't follow that 4^u must equal 4/3T. In fact, I think you'll find that 4^u is equal to T^(ln4/ln3). Your identity is only true when u=1, so it appears to give the solution in this case.

[@aayushshrestha1655]

X=2 2^2= 3^(2/2) +1

[@vladbuciu3099]

Damn man u do so much and i was needed only 30 sec to see x=2 and i watch to see if i was right.

[@SyberMath]

Wow! 😂

[@josevalencia9233]

Very elegant

[@SyberMath]

Thank you!

[@ocayaro]

At 2:31 you could have solved the problem in two more steps.

[@lagurim]

My Method: Replace x/2 by y Equation becomes: 2^(2y) - 3^y = 1 => (2^2)^y - 3^y = 1 => 4^y - 3^y = 1 => y =1 => x =2y = 2

[@saisuhas2969]

When I saw the problem I immediately know that x=2

[@Knowledge_Engineering]

Very easy problem, just take natural log both sides

[@rcnayak_58]

There is still a simpler and logical way to solve it. Let us say, x/2 = t. Then our equation will be 2^(2t) = 3^t + 1; which is 4^t = 3^t +1. This can be written as 4^t - 3^t =1. This will result only when when t = 1, that is 4 - 3 = 1. Any value of t more than 1 will cause more and more difference in their values only (>>1). Similarly any value of t > 0 and less than 1 will yield less than 1. The difference will be negative when t is negative. Therefore t = 1 is the only solution. So that our x becomes 2.

[@SyberMath]

Good thinking!

[@sitalroy_2462]

The most easy way to solve this is: Put the various value of x (x=1,2,3,......), and which value satisfy the equation ( such as L.H.S=R.H.S), and this x will be the solution of the equation. Here 2 is the solution.

[@SyberMath]

Only one solution?

[@sitalroy_2462]

@@SyberMath have u any doubt..?

[@ilaOzdemir]

(2^x - 1) =3 ^ (x/2) --> (2^x - 1)^2 = 3 ^ x . Then, lets say (2^x - 1) = u --> u^2 = 3^x ----> u= |3| and x= 2

[@altheakeithlynvlogz36]

1/2raised to x,,, (1/2) raised to x,,,,, are they equal?

[@rakenzarnsworld2]

x=2, 2^2=4, 3+1=4

[@qwadratix]

Hmm... The solution depends on spotting the accidental (or not-so accidental) coincidence in the given values leading to the cos/sin relation. I don't find that very satisfactory. What happens if you have a similar problem where the bases are not so convenient?

[@SyberMath]

Then you cannot find the exact answer

[@qwadratix]

@@SyberMath Only in the sense that it's irrational. Unless of course you get lucky again with something like, 2^x = 3^(x/2) -7 Solving the general case analytically would be a challenge I think. But then, I have trouble adding up my change in the local shop anyway.

[@dharmikdasari10]

It can be solved easily by trail and error method Let X=1 2¹ = 3½+1 condition fail Let X= 2 2²= 3+1 Ans: X=2

[@amehachewa4757]

How did you do get the number 2

[@SyberMath]

Guess and check

[@viktortodorov2109]

This is easier to take log with base 2. Then we will have linear equation for x. So easy can we do it

[@mdsahabuddin254]

You are wrong this is not that much easy as in the right hand side there are sum of two terms not multiplication.