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How to solve this "Harvard" entrance exam?

MindYourDecisions

29,424 views

Can you find all solutions to this popular puzzle?

I have re-uploaded a correction to the original video, which had a small typo of 9y instead of -9y (around 4:47 in How to solve this "Harvard" entrance exam?). Thanks everyone for catching that mistake!

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20 Comments

@kevinmartin7760 5 months ago

Once you allow complex solutions, each of the three "solutions" is actually an infinite family of solutions, spaced multiples of 2*(pi)(i) apart.
If you allow only real solutions, ln(3)/ln(2) is the only solution.

@TheeSleepWalker 5 months ago

Finally a problem I can solve on my own

@Ambientambiencechannel 5 months ago

That escalated fast.

@bryanalexander1839 5 months ago (edited)

What is meant by three roots? Three principal roots maybe, because otherwise there are infinitely many roots.

Just consider that the original equation is not a cubic equation, but rather has a variable for the exponent, so it makes sense that it will have infinitely many solutions.

For integer k,
x=[ln(3)+i2kπ]/ln(2),
x=[ln(–1.5±i0.5√31)+i2kπ]/ln(2)
gives the possible solutions.

@ccdsah 4 months ago

You could also use Cardano method. It's really easy since the equation is of this form

@athysw.e.9562 5 months ago

Man you upload so much videos recently. Hard to keep up

@savvasthoidis5403 4 months ago (edited)

Great video! Just a small parenthesis: while the cubic equation gives three values for y, only y=3 leads to a real solution for x, which is x = log_2(3). For the complex values of y, we would need to use the complex logarithm, but those wouldn't yield real exponents for 2^x anyway. The change of base formula is a great way to approximate the real root!

@XZeroOneArmour 4 months ago

whn bases are same , powers get added, as a novice 4x -->5.5( rough approx.) , x -->1.4 (approx) 😇

@silentbyte196 5 months ago

Real ones remember the og video

@KhamphacuocsongTV-CN4.0 5 months ago

Very good ❤❤😀

@MountainPieEnjoyer 4 months ago

its like between 1 and 2 i think

@HyeJin7771 5 months ago (edited)

I am here, feels like That sat trick guy. But still good video

@stvp68 5 months ago

What universities still have entrance exams? Does it mean just the SAT/ACT?

@Chandar-88 5 months ago

(2^×)^3+2^×=30
(2^×)^3+2^×-27-3=0
Let 2^×=a
a^3+a-3^3-3=0
a^3-3^3+a-3=0
(a-3)(a^2+3a+9)+a-3=0
(a-3)[(a^2+3a+9)+1=0
(a-3)=0
a=3
a^2+3a+10=0
(a+5)(a-2)=0
(a+5)=0
a=-5
(a-2)=0
a=2
Recall 2^×=a
2^×=3
×log2 =log 3
x=log2(3)//
2^×=2^1
×=1//
2^×=-5
×=In5+(2k+1)3.14i/In2//

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