Although the general formula for the cubic equation is off-putting and probably not worth it to learn, having a method to solve cubic equations can be useful, especially since questions like this are often cubic (because everyone knows the quadratic equation). To get from a general cubic ax^3+bx^2+cx+d=0 to a depressed cubic t^3+pt+q=0 we make the substitution x=t-b/3a and divide by a. To solve the depressed cubic you can make the substitution t=w-p/3w and multiply by w^3 to get a quadratic in w^3 : (w^3)^2+q(w^3)-p^3/27=0 which can be solved by the usual method. You only need one of the roots of the quadratic to get all 3 roots of the cubic. (Dividing by a or multiplying by w^3 are the more obvious steps and probably don't need to be remembered, the substitutions have similar structure but would probably still need to be remembered) Doing this only makes sense if you fail to guess the right solution (in this case 5+125=130 shouldn't be too hard, but if you are stuck, knowing there is a systematic way to solve this might be helpful)
An intermediate solution to solve a 3rd degree equation is to find an obvious solution n. It is then sufficient to develop (x−n)(ax²+bx+c) and to identify a,b and c with the coefficients of the polynomial.
faster way to prove x=log2(5) is the only solution: a function f(x) = 2^x + 8^x is the sum of 2 increasing functions (2^x and 8^x), that means f is also increasing. Since it's strictly increasing it's also injective, which means there can only be one real solution
2ˣ+8ˣ=130 f(x)=2ˣ and g(x)=8ˣ are strictly increasing so there is a unique solution for f(x)+g(x)=130 2ˣ+(2ˣ)³=130 2ˣ=X X+X³=130 X(1+X²)=130 X(1+X²)≈X³ and ∛130≈5.06 5(1+5×5)=5×26=130 X=5 2ˣ=5 x=ln5/ln2=log₂5
I dont get it, can't you just log both sides, take xlog2 + xlog8 = log130 So, xlog2 + 3xlog2 = log130 4xlog2 = log130, x=log130/log16, which gives 1.755? Please clear
When you take the log of a side of an equation you have to log the whole thing so it would be log(2^x +8^x) = log130 . Which isnt any easier to solve I think. This is shown by the fact that your answer log130/log16 doesn't equal his answer log5/log2.
Yes, as mentioned in the comment above, when you apply the log function to the left side of the equation, you cannot separate it into 2 different log equations. log(2^x + 8^x) does not equal to log(2^x) + log(8^x). Thanks for asking, and if it still not clear for you, I would be happy to try to provide more clarification!
@@MathMatrixUSA i think i understand, thankyou for clearing, but, what if we use variables instead of this, i.e, V= 2^x, log V = xlog2, similarly, logU= 3xlog2. So, LogU=3(LogV), therefore, U=3V, 130 = 4V, V = 32.5. now, V=2^x=32.5. Xlog2=log32.5, x=log32.5/log2 = 5.022?
Wait, I think I made a mistake while taking antilog, I think I should have taken antilog for 3 as well i suppose, so if I take 3 inside the log, it is just U=V^3, which we already know.. or is there any other way to take anti log without taking 3 inside? Wait, then, V+V^3= 130, V=5, so, 2^x=5, x=log5/log2, ahh, i fully understand now, thankyou so much for this video, or i wouldn't have understood this much about logarithms. (Can you still explain if there is any other way to anti log instead of taking 3 inside log itself?)
@@Vasantha-i3j I am very happy that you understand! Regarding your question, I believe that you need to apply log^-1 to the whole side of the equation to cancel out the Logs, so for when you wrote LogU = 3(LogV), that would be equal to log^-1(LogU) = log^-1(3(LogV)), and you can separate it into U = log^-1(3)*V, which is not U=3V, what you initially wrote. However, I am not completely sure, and I'm sorry if that sounds confusing, but you asked a really good question!
8^(x) does not equal 2^(3) * 2^(x). To get 2^(3) * 2^(x), it would need to be 2^(x + 3), but 8^(x) actually equals 2^(3x). When you put the final value this video shows into a calculator, you get ~2.32 instead of 3.85.
Although the general formula for the cubic equation is off-putting and probably not worth it to learn, having a method to solve cubic equations can be useful, especially since questions like this are often cubic (because everyone knows the quadratic equation).
To get from a general cubic ax^3+bx^2+cx+d=0 to a depressed cubic t^3+pt+q=0 we make the substitution x=t-b/3a and divide by a.
To solve the depressed cubic you can make the substitution t=w-p/3w and multiply by w^3 to get a quadratic in w^3 :
(w^3)^2+q(w^3)-p^3/27=0 which can be solved by the usual method. You only need one of the roots of the quadratic to get all 3 roots of the cubic.
(Dividing by a or multiplying by w^3 are the more obvious steps and probably don't need to be remembered, the substitutions have similar structure but would probably still need to be remembered)
Doing this only makes sense if you fail to guess the right solution (in this case 5+125=130 shouldn't be too hard, but if you are stuck, knowing there is a systematic way to solve this might be helpful)
Thank you so much for this insightful and helpful comment!
An intermediate solution to solve a 3rd degree equation is to find an obvious solution n.
It is then sufficient to develop (x−n)(ax²+bx+c) and to identify a,b and c with the coefficients of the polynomial.
faster way to prove x=log2(5) is the only solution:
a function f(x) = 2^x + 8^x is the sum of 2 increasing functions (2^x and 8^x), that means f is also increasing. Since it's strictly increasing it's also injective, which means there can only be one real solution
2ˣ+8ˣ=130
f(x)=2ˣ and g(x)=8ˣ are strictly increasing
so there is a unique solution for f(x)+g(x)=130
2ˣ+(2ˣ)³=130
2ˣ=X
X+X³=130
X(1+X²)=130
X(1+X²)≈X³ and ∛130≈5.06
5(1+5×5)=5×26=130
X=5
2ˣ=5
x=ln5/ln2=log₂5
Can't you just do 2^x+2^3x=130 and then go from there?
10x = 130
x = log 130/log10
x = 2.11394
10^2.11394 = 130
I dont get it, can't you just log both sides, take xlog2 + xlog8 = log130
So, xlog2 + 3xlog2 = log130
4xlog2 = log130, x=log130/log16, which gives 1.755? Please clear
When you take the log of a side of an equation you have to log the whole thing so it would be log(2^x +8^x) = log130 . Which isnt any easier to solve I think. This is shown by the fact that your answer log130/log16 doesn't equal his answer log5/log2.
Yes, as mentioned in the comment above, when you apply the log function to the left side of the equation, you cannot separate it into 2 different log equations.
log(2^x + 8^x) does not equal to log(2^x) + log(8^x). Thanks for asking, and if it still not clear for you, I would be happy to try to provide more clarification!
@@MathMatrixUSA i think i understand, thankyou for clearing, but, what if we use variables instead of this, i.e, V= 2^x, log V = xlog2, similarly, logU= 3xlog2. So, LogU=3(LogV), therefore, U=3V, 130 = 4V, V = 32.5. now, V=2^x=32.5.
Xlog2=log32.5, x=log32.5/log2 = 5.022?
Wait, I think I made a mistake while taking antilog, I think I should have taken antilog for 3 as well i suppose, so if I take 3 inside the log, it is just U=V^3, which we already know.. or is there any other way to take anti log without taking 3 inside?
Wait, then, V+V^3= 130, V=5, so, 2^x=5, x=log5/log2, ahh, i fully understand now, thankyou so much for this video, or i wouldn't have understood this much about logarithms. (Can you still explain if there is any other way to anti log instead of taking 3 inside log itself?)
@@Vasantha-i3j I am very happy that you understand! Regarding your question, I believe that you need to apply log^-1 to the whole side of the equation to cancel out the Logs, so for when you wrote LogU = 3(LogV), that would be equal to log^-1(LogU) = log^-1(3(LogV)), and you can separate it into U = log^-1(3)*V, which is not U=3V, what you initially wrote. However, I am not completely sure, and I'm sorry if that sounds confusing, but you asked a really good question!
why go through all the hassle, why can't you do, 2^x+2^3*2^x=130, which simplifes to 2^x(9)=130, which you just log it and get x = 3.85.....
8^(x) does not equal 2^(3) * 2^(x). To get 2^(3) * 2^(x), it would need to be 2^(x + 3), but 8^(x) actually equals 2^(3x). When you put the final value this video shows into a calculator, you get ~2.32 instead of 3.85.
You can see that x must be smaller than 3, because 2^3 + 8^3 is much greater than 130.