Solving a Pretty Exponential Equation in 2 ½ Ways

If you need to post a picture of your solution or idea:
#ChallengingMathProblems #exponentialequations

PLAYLISTS 🎵 :

[SyberMath]

Hello everyone, here is the graph that I forgot to show you in the video:

[carloshuertas]

Hey SyberMath! As soon as I see this math problem, I immediately know that x=1 in my head. But I actually did not know that there was another solution. Great explanation! Thanks a lot!

[moeberry]

You didn’t show us the graph lol you owe us Syber.

[misterdubity]

Can also take ln of both sides and then solve quadratic eqn by factoring.
(ln 2) (x^2 - 1) + (ln 3) (x - 1) = 0
(x - 1) * [(ln 2) (x +1) + (ln 3)] = 0
(x - 1) * [(ln 2) (x ) + (ln 2)+ln 3)] = 0
(x - 1) * ( x + (ln 6)/(ln 2) ) = 0

[stpetroff]

Всё, что нужно здесь для быстрого решения — это прологарифмировать обе части уравнения по основанию 2:
... | log2
log2 (2^x² * 3^x) = log2 (2 * 3)
x² + x*log2(3) = 1 + log2(3)
(x-1)(x+1 + log2(3)) = 0
Дальше очевидно.

[udpm]

I let m = log2(3), then 2^m = 3, and made them all the same base (2). After that I got a quadratic x^2 + mx - (m+1) = 0, which gives the same solutions as those in the video
I honestly didnt think about using these methods at all. Great video!

[threstytorres]

2^(x²) * 3^(x) = 6

Ln Both Sides:
Ln( 2^(x²) * 3^(x) ) = Ln(6)

Apply Log Rule: Ln(a*b) = Ln(a) + Ln(b)
Ln(2^(x²)) + Ln(3^x) = Ln(6)

Apply Log Rule Again; Ln(p^q) = qLn(p)
x²Ln(2) + xLn(3) = Ln(6)

Subtract Ln(6);
Ln(2)x² + Ln(3)x - Ln(6) = 0

Using the Quadratic Formula to get xwe get;
x= 1 and x= -Log(6)/Log(2)

[josephsilver]

Wow! I am impressed with method 1b and the third method. How do you have the time to consider and generate these multiple methods? I am quite happy to use ANY method to get the answer. Thanks for all of your hard work!

[Rbmukthegreat]

Was able to see the 1 from the thumbnail, but did not see those many beautiful methods! I has all my friends watch it and they also liked it! Wonderful video!

[flaviuzzz]

6:47 DUDE Why did no one ever tell me this ? THX! This is why I watch your videos!

[txikitofandango]

I used what you call method 1a, it was cool noticing that the discriminant was a perfect square

[sberacatalin]

Absolut, dar absolutely superb! TOT respectul! More! Next?!

[dublistoeo]

Awesome equation and methods! Thank you for the shout-out at the starting of the video. =)

[andylee]

Method 2 is pretty neat! Even if you didn't notice that x = 1 is a solution, once you had (x - 1) as an exponent on both sides you would naturally think to check whether x can be 1. So that would be yet another path to discovering x = 1 as a solution.

[solitudeineminor]

You're doing a really wonderful job brother

[thierryvieville]

Thanks again and again :) a nice challenge every day with smart and clever derivations.

Just for the pleasure to comment and … the pleasure to use log_2 :) If you log_2 both sides you get

x^2 + x l = 1 + l

writing l = log_2(3) and log_2(6) = log_2(2 3) = … and using some tiny algebra

which seems to mak visible that x = 1, thu sx^2 = 1 is a solution,

while we may further manipulate and derive

(x^2 - 1) + (x -1) l = (x - 1) (x + 1 + l) = 0

thus x = 1 or x = -1 - l = -1 - log_2(3)

nothing but a tiny variant of what is proposed in the video.

Take care.

[michaelempeigne]

i like the first method in video more since i naturally went that way.

[backwardsatom]

Nice, at (2^x)=(6^-1) I did log base 2 of 6^-1 = x, then change of base using ln

[eugenioaraya]

Hello SyberMath

Thank you for your always interesting videos. I would like to suggest to you to use subindexes whenever you have different solutions, as happens from the quadratic equation, using x sub1 and x sub2 and so on, so it is clearer to us the different solutions.

[walexandre]

My solution here I used base 3 instead of e. We get: log3(2)x²+x-log3(6)=0 => log3(2)x²+x-[log3(2)+1]=0. Solving the quadratic equation: x = [-1 +- Notice that 1+4log3(2)*(log3(2)+1) = (2log3(2)+1)². The solutions are x = 1 and x = log2(6).