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Hi Bobbym please help with this;
Find the value of x in the following equation
2^x - 2^-x = 2
This is similar to the one we solved previously, I tried and had [0] as an answer yet don't understand
The first one;
proceeding from the question:
3^2x 3^3 - 4 (3^x 3) + 1 = 0
= 3^2x 27 - 4(3^x 3) + 1 = 0
= (3^x)^2 27 - 4(M 3) + 1 = 0
Let 3^x = M
(3^x)^2 27 - 4 (M 3) + 1 = 0
= 27M^2 - 12M + 1 = 0
= 27M^2 - 9M - 3M + 1 = 0
= 9M(3M - 1) - 1(3M - 1)= 0
= (9M - 1) (3M - 1) = 0
Let M = 3^X
9M - 1 = 0
M = 1/9
3^X = 1/9 = 3^ -2 = -2
X = -2
3M - 1 = 0
M = 1/3
3^X = 1/3 = 3^-1 = -1
{ X : X = -1, -2}
Actually with respect to the first one I did used factorization in the course of manipulation, see;
4 3^(2x+1) + 17 3^X -7 = 0
= 4 (3^x)^2 3^1 + 17 (3^x) - 7 = 0
Let 3^x = M
= 4 M^2 3 + 17M - 7 = 0
= 12M^2 - 4M + 21M - 7= 0
= 4M(3M - 1) + 7(3M - 1) = 0
=(4M + 7) (3M - 1) = 0
Let M = 3^x.
4M + 7 = 0
= 4M = -7
=M = -7/4
Again,
3M - 1 = 0
3M = 1
M = 1/3 Let 3^x = M
3^x = 1/3
3^x= 3^-1.
x = -1
I used the same method to do the first one
Hi;
I have endeavored to solve them, I literally used hours, which I had {-1, -2} for the first one and {-1} for the second problem. At the back of the book it has {-1} as the answer for the second problem, so I think I am correct.
Did you say you used factorization and quadratic to solve them, and what are your answers?
Thanks
At the back of the book it gave these as it answers; {x : x = -1, -2}
I will try to solve, I doubt if will get those answers.
Thank you Bobbym
By the way, if you can supply me with a link that teaches or deals with polynomial happy will I be
Besides, those problems are treated under indices
Thanks
Polynomials are in the book, but I am yet to learn them properly.
In truth, they are to be solved separately. The book has solved similar problems, but separately, in fact they are separate problems but those ones are difficult for me to solve that is why I posted them at the same time.
Please look at #307 it is numbered like that, with a heading words as;
"Find the truth set of the equations"
Ok I have now gotten what you mean,it is that both one and two should be solved separately
No not seperately, but simultaneously
Anoni, you're right Thank you
Good!
It is ( 3^(x+1) ) the 3^(x+1) are in a bracket
There is no bracket around 2x+3 in the book, but I will puts them for simplicity sake
Again,
3^(2x+3) - 4(3^(x+1)) + 1 = 0
It is not from a book I made it up , all the same I have understood you, see these;
Find the truth set of the equations
(1) 3^2x+3 - 4(3^x+1)+1 = 0
(2) 4( 3^(2x+1) ) +17( 3^x ) - 7 = 0
It is numbered like this
I am subtracting x^2 from x^4
Please help me here;
(X^2 - X^4) =
What would be the answer?
I have now learnt it, thank you bobbym
I have a couple of problems I will post.
Why did you not say x(y^1/2) = xy^3/2 I think the x has 1 as its power, please I am not getting you with this x(y^1/2) = xy^1/2. Please explicate.
So, if there was X in the bracket having 1/2 as it power how would the answer be? like x(y^1/2 - x^1/2)
Thank you.
I was talking of the X which is outside the bracket. Look at a similar case at #182 line one, after you had multiplied 2^-x through, the 2 at right side became 2^x+1, this indicates that the 2 already had 1 as its power, and in this problem I perceive that the X has 1 as its power though not visibly expressed or written , what do you say?
Thanks
To begin with I would say the X instrinsically raised to the power positive one[1], so I will say 1+1/2 = 3/2. Therefore;
XY^3/2
I have done that, still I only arrive at the same answer.