A Homemade Exponential Equation | 3^{x^2}9^x=27^{{sqrt(x)}

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[walterufsc]

3^x² . 9^x = 27^√x
3^x² . 3^2x = 3^3√x
3^(x² + 2x) = 3^3√x ==> x² + 2x = 3√x
Squaring both sides: x^4 + 4x^3 + 4x² = 9x
x^4 + 4x^3 + 4x² - 9x = 0
x(x^3 + 4x² + 4x - 9) = 0 ==> x = 0 (1st solution)
x^3 + 4x² + 4x - 9 = 0 ==> x = 1 (2nd solution)
By using polynomial division:
(x^3 + 4x² + 4x - 9)/(x - 1) = x² + 5x + 9
x² + 5x + 9 = 0 ==> x = (-5 ± i√11)/2 (two complex solutions)

[ytlongbeach]

thx for including the complex solutions.

[chaosredefined]

When determining if u=1 is a solution to (u^3 + 2u - 3), you can split the terms into odd and even powers, i.e. (u^3 + 2u) and (-3), then apply it. Then apply the u=1 case giving you (3) and (-3). As they are equal and opposite, this means that u=1 is a solution. However, this has the added advantage that, if they are just plain equal, then the negative of the value you tried (u=-1) is a solution.

[popitripodi]

This was an easy one!!! I enjoyed it!!! I like that you always show the graph!!!❤❤❤❤

[MichaelJamesActually]

Would you pls do a video on how to work thru determining the remaining terms when x-1 is a factor? I struggle thru it, and think I have it down. would be good to go thru a few examples from quadratic up to quintic. thanks!

[christopherellis]

3 times 9 equals 27 was more than a hint

[jandirpassos]

You can find out the complex and real roots by Briot-Ruffini algorithm, too.

[damiennortier]

This is the first day you make 2 video and I love it 💗💗💗

[אלדד_אלדד]

S△ABC=(AB²+AC²)/4, find the angles of the △.

[abdulwasaykamran]

After seeing your equation I was able to get the value of x=1 but I wasn't able to answer other values of x cause I only read in class10 this is why I don't know other process. My other equation that I got is x^2+5x+9 which don't go middle term break or I can't use the formula (-b+-√b^2-4ac) ÷2a.

[andrasnoll]

Ok I couldn't figure out the cimplex in head, but first glance anyone could tell 0 and 1 i mean 3*9=27 so (3^1)*(9^1)=(27^1) so since both √1 and 1^2 is just 1
(or -1 but that doest work in this praticilural contast)
and since any nomber ^0 is 1 and 1*1=1 than by extension
(3^0)*(9^0)=(27^0) also works 'cause 0^2 and √0 are also 0

[yoav]

Nice!! I tried this eq x^2+2x=3sqrtx over the complex and wolfram gives only x=0, 1 as the solutions. I wonder why, maybe you have an idea. 💯💯

[endercool]

I m not sure you are allowed to square both sides on the last or even the last two solutions though because you have the equation sqrt(x) = [(- 1-sqrt(11)i]/2 wich is negative and one rule is that tou can never have the square root of a number x be equal to something negative, cirrect if im wrong, but thats also the case in the complex world, while you might be able to solve for x^2=-1 and say that x=i for example, you can never say that because sqrt(x) =-1---> x^2=1 so x=+/-1 because thats a fake answer

[albenismiranda]

It would have been more interesting if at the end you had graphed the two initial functions and not those of the exponents when doing the equalization of bases. Greetings

[kianmath]

real sols: x=0, 1' imaginary: (-1+-isqrt11)/2

[dentonyoung]

Interesting that an exponential equation in real numbers would have complex solutions....

[broytingaravsol]

still nothing i can reach out if x would be lie in a complex domain for such a real pattern