A Fun Exponential Equation

Рет қаралды 3,643

Күн бұрын

Пікірлер: 18

@hallfiry Ай бұрын

More streamlined start: pull a 15 from the denominator to get a clean 15^x down there. Then divide through, so you get (9/15)^x and (25/15)^x. Simplify those fractions and set z=(5/3)^x, so you get 15(z-1/z)=16. Then solve as usual.

@DrBarker Ай бұрын

I like this as a starting approach. When you see the z - 1/z term, it starts to hint at the possible use of complex numbers later on, and reminded me of Vieta's substitution.

@RGP_Maths Ай бұрын

From thumbnail: let a=3^x, b=5^x So (a²-b²)/(ab/15)=16 So 15(a²-b²)=16ab So 15a² - 16ab -15b² =0 So (3a - 5b)(5a + 3b) = 0 But a&b are both positive, therefore 3a=5b So a/b = 5/3 So (3/5)^x = 5/3 So x = -1. I didn't consider non-real x though!

@thorvalddox7715 Ай бұрын

Aren't you missing some complex solutions because (3/5)^(x-1) = (5/3)^2 also gives rise to a set of complex solutions in addition to the real one?

@DrBarker Ай бұрын

Yes, very good point! It feels a bit unnatural given that there is a real solution to (3/5)^{x-1} = 25/9, but we can apply the same method as we did for (3/5)^{x-1} = -1 to get another family of complex solutions to (3/5)^{x-1} = 25/9.

@General12th Ай бұрын

Hi Dr. Barker!

@ronbannon Ай бұрын

Thanks for sharing. Very nice treatment and at a level that most will understand. You're an excellent teacher.

@DrBarker Ай бұрын

Thank you!

@garkechify Ай бұрын

If you consider complex solutions, the first case (u=25/9) yields to multiple values of x, too: x = -1 + 2k \pi i / ln(3/5)

@DrBarker Ай бұрын

Very well-spotted, and neat how this still captures the real solution when k = 0.

@mathmachine4266 Ай бұрын

a=3^x, b=5^x (a²-b²)/(ab/15)=16 15(a²-b²)=16ab 15a²-16ab-15b²=0 a=(16b±√(256b²+4*225b²))/30=(8b±√(64b²+225b²))/15=b(8±√(289))/15=b(8±17)/15 a is either 25b/15=5b/3, or -9b/15=-3b/5 First case: 3^x=5*5^x/3, 3^(x+1)=5^(x+1) (x+1)ln(3)=(x+1)ln(5)+2πi*N, where N is some integer (x+1)ln(3/5)=2πNi x+1=2πNi/ln(3/5)=2πNi/ln(5/3) x=2πNi/ln(5/3)-1 Second case: 3^x=-3*5^x/5, 3^(x-1)=-5^(x-1) (x-1)ln(3)=(x-1)ln(5)+(2N+1)πi, where N is some integer (x-1)ln(3/5)=(2N+1)πi x-1=(2N+1)πi/ln(3/5)=(2N+1)πi/ln(5/3) x=(2N+1)πi/ln(5/3)+1 Assuming N is any integer, x=-1+2Nπi/ln(5/3) or x=1+(2N+1)πi/ln(5/3) That is to say, x is either -1 plus an even multiple of πi/ln(5/3), or +1 plus an odd multiple of πi/ln(5/3). And the only real solution is x=-1. The reason I got more solutions is because (3/5)^(x-1)=(3/5)^-2 does not necessarily imply x-1=-2.

@ProactiveYellow Ай бұрын

Reduce down and separate the fraction difference into 3^{x+1}/5^{x-1}-5^{x+1}/3^{x-1}=16, factor a 3/5^{-1}=15 out of both and divide through for (3/5)^x-(3/5)^{-x}=16/15. I like this, nice symmetry here. Divide by 2 and make the substitution 3/5=e^{ln3/5} to get (e^{xln⅗}-e^{-xln⅗})/2=8/15, but that's just sinh(xln⅗)=8/15. Unwrapping x gets us that x=arsinh(8/15)/ln(3/5). Recall that the logarithmic form of arsinh(k)=ln(k+√(k²+1)), thus arsinh(8/15)=ln(8/15+√(64/225+1))=ln(8/15+√(289/225))=ln(8/15+17/15)=ln(5/3), so x=ln(5/3)/ln(3/5)=-ln(⅗)/ln(⅗)=-1 Therefore the real solution -1 is a solution to the equation, and because sinh is a bijective function on the reals, arsinh is as well, thus -1 is the unique real solution.

@alipourzand6499 Ай бұрын

The pitagorian triplet 3, 4, 5 is showing up! Great video.

@LukeFlavel Ай бұрын

It's a weird feeling to have the maths knowledge to interpret what you're saying but not the capability to solve a question like that in my own right 😅 Very entertaining though, I did not expect things to get so deep

@alaingamache3908 Ай бұрын

Spoilers below !!

@lloydgabrielburden-garabed5221 Ай бұрын

For an elegant solution to find the real value x=-1 satisfying the equation you can use exponent rules to quickly manipulate the equation into one which gives you the answer for free: Use power of a power rule in numerator and product rule in denominator: [3^(2x)-5^(2x)]/[3^(x-1)×5^(x-1)]=16 Split the fraction and cancel like terms: 3^(x+1)/5^(x-1) - 5^(x+1)/3^(x-1)=16 Multiplying first term by 5^2/5^2 and second term by 3^2/3^2, use product rule to only have powers of x+1 and quotient rule to combine terms: 25(3/5)^(x+1)-9(5/3)^(x+1)=16 From there it's obvious that because 25 -9=16 the only solution is when the both exponential terms are equal to 1, i.e. when the exponent is 0, or when x=-1

@MyOneFiftiethOfADollar Ай бұрын

Excellent, who would have thought original was in quadratic form?!?!! Important to pick the constant(16 in this case) such that you get factorization over the rationals.