A Nice Exponential Equation | Math Competitions

[@WahranRai]

7:48 Let f(t) = (16/41)^t + (25/41)^t - 1 with t = x/2 f(t) is *injective* and f(1) = 0 ---> x = 2 is the unique solution of the proposed equation

[@Blaqjaqshellaq]

If f(t)=(16/41)^t + (25/41)^t, then f'(t)=(ln16-ln41)*(16/41)^t + (ln25-ln41)*(25/41)^t, which is negative for every real value of t, therefore f(1) is a singular value.

[@dwm1943]

To find x = 2 as a solution, about 5 seconds (I am having a slow day). To prove this is a unique solution, much harder, and I had no idea how to do it. So, once again, thank you. Neat.

[@AllanPoeLover]

Should X < 0 has posibility to find a solution?

[@carly09et]

I started by assuming a solution exists, then 'guess and check' to see. First guess of x=2 worked. So consider the general solution space.

[@ruud9767]

This video would have been shorter without the preamble and digressions. Plot 2^2x + 5^x - 41^(x/2). Discover that x=2 is a solution. Argue why it is the only solution. Done.

[@dapulse7147]

How is this solved without guessing?

[@MrMatteuccio82]

x=2 after 3 seconds, 16+25=41

[@thexavier666]

The t method is quite nice

[@giuseppemalaguti435]

x=2..x√41^x..x>2 il contrario,immagino lavorando sulle derivate

[@edcoad4930]

Just be inspection, x must be 2.

[@Quest3669]

25+16= 41

[@archangecamilien1879]

x=2, perhaps, lol...

[@scottleung9587]

I got x=2 by inspection.

[@yakupbuyankara5903]

X=2

[@MatheusSouza-lw9wv]

bom, assim, não sou gênio nem nada mas é meio evidente que x=2, porque temos 2^2x= 4^x, assim ficamos com: 4^x + 5^x = 41^[x/2] quando x=2, temos: 16+25=41, o que é verdadeiro. Sinceramente, acho essa muito fraca.

[@DJ.Nihad6174]

Cool,cool😂😂

[@Don-Ensley]

Excellent f(t) method showing x= 2 is the only solution!! I did it a little differently and not as effectively as you. problem 2²ˣ + 5ˣ = 41ˣᐟ² Notice that 16 + 25 = 41 which fact is most useful for this problem. Since 2⁴ = 16, and 5² = 25, we have 2⁴ + 5² = 41¹ as a consequence of simple arithmetic. Setting 2x = 4, x = 2, x/2 = 1 perfectly fits. Let f(x) = 4ˣ+ 5ˣ f'(x) = 4ˣln 4 + 5ˣ ln 5 > 0 for all x, so it is increasing continuously with increasing x. f"(x) = 4ˣ(ln 4)² + 5ˣ(ln 5)² > 0 for all x so f opens upward and has no inflection point. lim f(x) = 0 so the x-axis is a vertical x →- ∞ asymptote in the negative x direction. and let g(x) = 41ˣᐟ² g'(x) = (1/2) 41ˣᐟ² ln 41 > 0 for all x, so g is continuously increasing . g"(x) = (1/4) 41ˣᐟ² ( ln 41)² > 0 for all x so g opens upward and has no inflection point. lim g(x) = 0 so the x-axis is a vertical x →- ∞ asymptote in the negative x direction. In the range [0, 2], f(x) and g(x) vary between f(0) = 2 g(0) = 1 so g starts out less than f At x = 1, f(1) = 9 g(1) = √41 g is obviously still less than f with √41 between 6 and 7. and f(2) = 41 g(2) = 41 where f and g are equal. g now grows at a faster rate than f since g'(2) = (1/2) 41 ln 41 > 16 ln 4 + 25 ln 5 For x > 1.39844 approximately, g grows faster than f. Therefore g will always be greater than f for x > 2, therefore x = 2 is the only solution. answer x = 2

[@rakenzarnsworld2]

x = 2