A Nice Exponential Equation | Math Olympiads

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I saw this question from Dr PK Math. You two are great

Min-cvnt

I think I've seen this problem before, but I wasn't able to solve it until now. Thanks so much for presenting it again!

scottleung

It's funny how range and domain issues become more relevant in higher level math problems.

spelunkerd

I like very much this kind of problem. It can always be transformed in a fractional polynomial. The general analytical solution for the equation a^x + b^x = c^x can be found in the paper (only two pages) "On the Solutions of a^x + b^x = c^x", available for download on researchgate. For example, if one has 6.1^x + 4.1^x = 9.1^x the solution would be 1.2053 (up to four digits). The solution of the problem described in the video is 1.1868 (up to four digits).

rubensramos

When one has a^x + b^x = c^x and a, b and c do not have primes in common, the Golden ration does not appear. For example: 7^x + 5^x = 11^x. In this case the solution is 1.1474 (up to 4 digits).

rubensramos

Also we can divide by (2^x)(2^x), solution could also be written as log (base = 3/2) of (1+sqrt(5))/2

creounity

You assumed that x was real here, but there are of course the complex solutions that make (2/3)^x negative:

9ˣ = 4ˣ + 6ˣ : let u = (3/2)ˣ and divide through by 4ˣ
u² = 1+u: this gives solutions
u = phi and u = -1/phi

Now x = log(u) / log(3/2) gives two families of solutions:

u = phi -> x = log(phi) / log(3/2)
u = -phi/2 -> x = -log(phi) / log(3/2) + i*pi*(1+2n) / log(3/2)

This is using log(-x) = log(x) + i*pi*(1+2n) for x > 0 and n any integer.

blackraven

x =1.1869
9^x /6^x = 4^x/6^x +6^x/6^x (divide b oth sides by 6^x)
(3/2)^x = ( 2/3)^x + 1
let p = (3/2)^x, then
(2/3)^x = 1/p ; hence
p=1/p + 1
p^2 = 1 + p multiply both sides by p
p^2 -p-1 =0
Using quadratic formulae gives 1/2 + sqrt 5/2 or (1 + sqrt 5)/2 or 1.61803
hence (3/2)^x = 1.61803
x log 3/2 = log161803 or ( 1+ sqrt 5)/2
x= 0.2089 / 0.176
x = 1.1869

devondevon

Technically there are two solutions Syber. If you consider the complex one. If you take ln of both sides of the negative quantity you obtain ln((-1-sqrt(5))/2) which can be written as i*pi*ln(phi). And there are infinitely many solutions if you just keep adding 2pin within the i*pi product. Where n is an integer and phi is the Golden Ratio.

moeberry

The natural logarithm of a negative number is indeed undefined in the real domain, but is is defined in the complex domain. ln(x) where x < 0 = ln|x| + πi

JamesDavy

x can also be presented as [ln(5^1/2-1)] - ln 2]/[ln(2)-ln(3)]

Blaqjaqshellaq

Can someone help me with some maths? So I asked chatgpt and it told me this:

My question:

A Vaccine has a of 100. The lethality of the illness is 0, 00002%. How many persons do you have to vaccinate to prevent one deadly case of the illness?

Answer:

To calculate the number of persons that one would have to vaccinate to prevent one deadly case of the illness, you would need to divide 1 by the (100) and then divide that by the lethality rate of the illness (0.00002%).

1/(100*0.00002) = 5, 000, 000

So, one would have to vaccinate about 5, 000, 000 persons to prevent one deadly case of the illness.
Keep in mind that these are hypothetical numbers and the actual numbers may vary depending on the vaccine and the illness. Also, this is based on the assumption that the vaccine is 100% effective which is not always the case.

But is this really true? Isnt 500.000.000 the correct answer? If it prevents 1 case in 100 injects, and every 5.000.000 cases 1 death occurs (100/0, 00002), then you would have to vaccinate 500.000.000 persons to prevent 1 death with the vaccine.

Can you help me with whos right here?

systemabc