Mexico - A Nice Math Olympiad Exponential Problem

Maths Olympiads are held all around the world to recognise students who excel in maths. The test is offered at many grade levels and provides them with numerous possibilities to win certifications, awards, and even scholarships for higher studies.

[honeytgb]

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

[logminusone]

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.

[andreadanieli]

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)

[georiashang]

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.

[shannoo]

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.

[mangofan]

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

[PlasmaFuzer]

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

[josephmiller]

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.

[wiseview]

All these problems seem a little too easy for olympiads

[arcticwolf]

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

[onetoo]

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 .

[dimchodimov]

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.

[akshatgour]

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 😅

[kathrynstemler]

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.

[IoT_]

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.

[schlingel]

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.

[IvyANguyen]

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.

[kanguru_]

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.

[hardtimes]

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).

[kevinsolari]

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<5 to small.