Why you log both sides, when you got 9^(x-1)=1/4? You can use exponent rule and get (9^x)* (9^-1) =1/4, so (9^x)*(1/9)=1/4. Then multiply both sides by 9 and get 9^x=9/4. Now you can use the definition of logarythm and get x=log 9 (9/4)=log 3 (3/2)
Куча не нужных действий По свойствам степени а^х/а^у=а^(х-у) Выносим за скобки 9^(х-1) Тогда по свойствам степени 9^(х-1)*(9^2-1)=20 9^(х-1)=1/4 Если учесть что 9 =3^2, 4=2^2 То 3^(х-1)=1/2 Логарифмировать по основанию 3 х-1= log(3)(1/2) x=1-log(3)2 2 минуты, а не позор на 10 минут Зачем логарифмировать по неизвестному основанию?
Curious why you don't just take log base nine of both sides? Is there an advantage to going through common log first and using the change of base formula?
I noticed that he simplified the answer to 1-(log2)base3, whereas leaving the answer at 1-log4 base 9 would make it easier to check the answer since the coefficients of the original equation are both 9
Vamos a sacar el número atómico que es el Calcio (Ca) que es el elemento de la fusión de protones que tiene una estrella en sus gases como materia de la formación de materiales químicos de ebullición. 9^⁽*⁺¹⁾ - 9^⁽*⁻¹⁾ = 20 p = 9^⁽*⁾ 9.p - 9^⁽⁺¹⁾ . p = 20 9.9^⁽*⁾ - 1/9 . 9^⁽*⁾ = 20 81^⁽*⁾ - 1/9 . 9^⁽*⁾ = 20 80^⁽*⁾/9 . 9^⁽*⁾ = 20 9^⁽*⁾ = 20.9/80 9^⁽*⁾ = 9/4 x = ₉ 9/4 x = Log₃(3/2)
thanks you. good exelent!!!!!!
81*9^x-9^x=180 , 9^x=9/4 , x=log(9/4)/log9 , test , OK ,
My way is ... different (few steps):
9^(x + 1) - 9^(x - 1) = 20
9ˣ·9 - 9ˣ/9 = 20
9ˣ·81/9 - 9ˣ/9 = 20
(9ˣ·81 - 9ˣ)/9 = 20
9ˣ·81 - 9ˣ = 180
9ˣ·(81 - 1) = 180
9ˣ·80 = 180
9ˣ = 9/4
x = ln(9/4)/ln(9)
x ≈ 0.369070
---
/// final result:
■ x = ln(9/4)/ln(9) ≈ 0.369070
🙂
This is the solution I found. Quicker and easier.
y = 9^x
9y - y/9 = 20
y(80/9) = 20
y = 9^x = 20 ÷ (80/9) = 9/4
2xlog3 = 2(log3 - log2)
x = 1 - (log2)/(log3)
ANS x = 1 - log2/log3 = 0.369
Why you log both sides, when you got 9^(x-1)=1/4? You can use exponent rule and get (9^x)* (9^-1) =1/4, so (9^x)*(1/9)=1/4. Then multiply both sides by 9 and get 9^x=9/4. Now you can use the definition of logarythm and get x=log 9 (9/4)=log 3 (3/2)
Куча не нужных действий
По свойствам степени
а^х/а^у=а^(х-у)
Выносим за скобки 9^(х-1)
Тогда по свойствам степени
9^(х-1)*(9^2-1)=20
9^(х-1)=1/4
Если учесть что 9 =3^2, 4=2^2
То
3^(х-1)=1/2
Логарифмировать по основанию 3
х-1= log(3)(1/2)
x=1-log(3)2
2 минуты, а не позор на 10 минут
Зачем логарифмировать по неизвестному основанию?
Ну мне все-таки странно видеть и логарифмы по тройке. Сразу стандартно свести к -ln2/ln3, потом оформить. И это задача типичная, не олимпиадная.
I just took logs base 9 and got x = Log 'base 9' of (2.25). It checks out.
9^(x+1)-9(x-1)=20
9*9^x-(9^x)/9=20
9^x*(9-1/9)=20
9^x*(81-1)/9=20
9^x*(80/9)=20
9^x=20*( 9/80)=9/4j
X=(log (9/4)
9
=log 9- log 4
9 9
= 1-log 4
9
,
Yes. My approach also. This seems to be the most straight forward and simplified method taking 30 seconds not a 5 minute scenic tour!
Solution:
9^(x+1)+9^(x-1) = 20 |*9 ⟹
81*9^x+9^x = 180 ⟹
82*9^x = 180 |/82 ⟹
9^x = 180/82 = 90/41 |ln() ⟹
x*ln(9) = ln(90/41) |/ln(9) ⟹
x = ln(90/41)/ln(9) ≈ 0.3578
It's very cool how the - magically changed to +
Curious why you don't just take log base nine of both sides? Is there an advantage to going through common log first and using the change of base formula?
I noticed that he simplified the answer to 1-(log2)base3, whereas leaving the answer at 1-log4 base 9 would make it easier to check the answer since the coefficients of the original equation are both 9
What country is the music?
Vamos a sacar el número atómico que es el Calcio (Ca) que es el elemento de la fusión de protones que tiene una estrella en sus gases como materia de la formación de materiales químicos de ebullición.
9^⁽*⁺¹⁾ - 9^⁽*⁻¹⁾ = 20
p = 9^⁽*⁾
9.p - 9^⁽⁺¹⁾ . p = 20
9.9^⁽*⁾ - 1/9 . 9^⁽*⁾ = 20
81^⁽*⁾ - 1/9 . 9^⁽*⁾ = 20
80^⁽*⁾/9 . 9^⁽*⁾ = 20
9^⁽*⁾ = 20.9/80
9^⁽*⁾ = 9/4
x = ₉ 9/4
x = Log₃(3/2)
Writing out is too slow!!!