Divide both sides by 16^(1/x) [(3/2)^1/x]^2 +[3/2)^1/x-1=0 Take (3/2)^1/x = a a ^2 + a -1=0 >(a +1/2)^2 -5/4=0 > a+1/2= +/- √5/2 then go on It will save time
To start with, divide both sides by the right hand side. This gives us x root of 9/4 + x root of 3/2 = 1. For simplicity, let's denote by y the quantity x root of 3/2. Then we have y²+y=1. Thus y = (square root of 5 -1)/2. And x = log base y of 3/2. This is the same as log base e of 3/2 divided by log base e of y. Evaluating this numerically yields x = -0.84259... My five math books for middle and high school show the deep simplicity of mathematics.
The lesson was familiarity with exponents and log of numbers. So appropriate length and careful explanation. So nice introductory lesson, piece by piece
[9^(1/x)]+[6^(1/x)]=4^(1/x) [(9/4)^(1/x)]+[(3/2)^(1/x)]=1 Let u=(3/2)^(1/x)>0 u²+u-1=0 --> u=½[-1±sqrt(5)] =-1±½[1±sqrt(5)] Note that ½[1+sqrt(5)]=ß, golden ratio ½[1-sqrt(5)]=-1/ß Note that the 2nd root of u
Why did you use a calculator at the end? If a calculator was allowed, you could have used it earlier. I just do not understand the logic behind this process.
Divide both sides by 16^(1/x)
[(3/2)^1/x]^2 +[3/2)^1/x-1=0
Take (3/2)^1/x = a
a ^2 + a -1=0
>(a +1/2)^2 -5/4=0
> a+1/2= +/- √5/2
then go on
It will save time
Excellent 👌
To start with, divide both sides by the right hand side. This gives us x root of 9/4 + x root of 3/2 = 1. For simplicity, let's denote by y the quantity x root of 3/2. Then we have y²+y=1. Thus y = (square root of 5 -1)/2. And x = log base y of 3/2. This is the same as log base e of 3/2 divided by log base e of y. Evaluating this numerically yields x = -0.84259... My five math books for middle and high school show the deep simplicity of mathematics.
Hi Andre, pleasse could you let me know the titles and authors of your school books.
The lesson was familiarity with exponents and log of numbers. So appropriate length and careful explanation. So nice introductory lesson, piece by piece
Thanks 😊
Решаем методом устного счета.
Сокращаем уравнение на 36^х.^1/2
(2/3)^x^1/2=y
y^2-y-1=0
y=(1+√5)/2
y=(1-√5)/2
х^1/2=log 1,6
x^1/2=log(-0,6)
Good luck!
[9^(1/x)]+[6^(1/x)]=4^(1/x)
[(9/4)^(1/x)]+[(3/2)^(1/x)]=1
Let u=(3/2)^(1/x)>0
u²+u-1=0 --> u=½[-1±sqrt(5)]
=-1±½[1±sqrt(5)]
Note that
½[1+sqrt(5)]=ß, golden ratio
½[1-sqrt(5)]=-1/ß
Note that the 2nd root of u
Спасибо, вы показали решение сложного 😮
You are welcome 😊.
I am glad to hear that.
36 1/x+ 241/x = 16 1/x 4 1/x * 9 1/x + 4 1/x * 6 1/x =4 1/x * 4 1/x
4 1/x (9 1/x + 6 1/x) =4 1/x * 4 1/x 9 1/x + 6 1/x = 4 1/x
1/x log 9 +1/x log 6 = 1/x log 4 1/x( log 9 + log 6) = 1/x log 4
Fo above to be true log 9 + log 6= log 4
That statementwould never be true. Therefore there is no real value for x to fulfil the above equation.
Why did you divide them by 36
To simplify and if you are wondering why not any other number you can but he didn't do it because you need only one answer
14:35
Любой инженер решит это уравнение за 5 минут с калькулятором, который имеет функцию взять корень любой степени
Good account but please take away those appalling reminders when you change the page. That bell is enough to you crazy.
Noted!
Why did you use a calculator at the end? If a calculator was allowed, you could have used it earlier. I just do not understand the logic behind this process.
It’s fine. He could leave the answer with logs
@asqar_math however, for the residual logs a calculator was necessary.
(3/2)^(2*(1/x))+ (3/2)^(1/x)-1=0 , let u=(3/2)^(1/x) , u^2+u-1=0 , u=(-1+/-V(1+4))/2 , u=(-1+/-V5)/2 , (3/2)^(1/x)=(-1+V5)/2 ,
1/x=log((-1+V5)/2)/log(3/2) , x=log(3/2)/log((-1+V5)/2) , x=~ -0.842592 , 1/x=~ -1.18681 ,
test , 36^(1/x)+24^(1/x)=~ 0.37233 , 16^(1/x)=~ 0.37233 , same , OK ,
Great job 👏
@@SpencersAcademy Thanks!
@@SpencersAcademy Thanks!
36^(1/x) + 24^(1/x) = 16^(1/x)
9^(1/x) + 6^(1/x) = 4^(1/x)
9^(1/x) + 6^(1/x) - 4^(1/x) = 0
(3/2)^(1/x) = a
a^2 + a - 1 = 0
a = (-1 +/- √5)/2
1/x = ln((-1 +/- √5)/2) ÷ ln(3/2)
x = 1/[ln((-1 +/- √5)/2) ÷ ln(3/2)]
Very difficult