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I have got it now! thank you
Thank you Bobby, I have digested that. But I am wondering why you turned the 2/3 on the 16 to be 3/2. I was thinking it should be; x = 16^2/3
What do you say?
Because I latexed it up incorrectly!
Originally, the power was 2/3, how did it became 3/2 as in 16^3/2?
Divide both sides by 5.
Raise both sides to the 3 / 2 power > [why didn't you say 2/3 power?].
Please, could you break it down a little bit for me to know how 16^2/3 finally gave you 64?
Thank you Bobbym.
Last but not the least, Please, I want you to clearly solve this apparent headache for me;
Solve for X.
5X^2/3 = 80
Thank you Bobbym!
Is the negative one answer correct?
Please confirm
The book has only 2 as the answer
This is my factorization
m^2 - 2m - 8 = 0
m^2 - 4m + 2m - 8 =0
m(m -4) 2(m - 4) = 0
(m + 2)(m - 4) = 0
let m = 2^x
m + 2 = 0
m = -2
2^x = -2 = -1 is this answer correct?
m -4 = 0
m = 4
2^x = 4 = 2^2 = 2
And are you saying it is wrong?
I am wondering if that negative one [-1] is correct.
I have done it, but this is a part of mine
(m +2) (m - 4) = 0
Let m = 2^x)
m = -2
2^x = -2 = -1 is this correct?
m = 4
2^x = 2^2 = 2
In fact, this is the original question;
Find the value of X in following equation.
2^(2x-3) - 2^(x-2) - 1 = 0
Again,
2^2x - 2 . 2^x - 8 = 0
I had 1 and 2 as answers, but the book has only 2 as the answer.
Errr, I see, I miscalculated it. Both 2 and 3 when plug in give 13
Thank you Bobbym!
These are the possible answers to the question; A. 13 B. 19 C. 25 .D 37
If X + 6/X = 5 find the value of X^2 + 36/X^2
I had 10 and 13 as the answers but I am wondering as to why it has only 13 in the possible answers, can you explain why?
I had 2.4142 and -0.4142 as the answers, is these correct?
Please confirm.
The y's in post #352 are changed to 2^x and solved for.
I guess if one use 2^x still factorisation is impossible unless the quadratic formula. What do you say?
THANK YOU BOBBYM, GOD BLESS!!!
I did not put the question well, I meant the answers will be multiple answers
Thank you very much Bobbym! I will post another perhaps tommorow, finally finally if so then, the answers must be two as you have shown above
Please I think these are correct,
a = 1, b = -2, c = -1 Before you will plug them in.
I was thinking of using the formula; -b positive or negative, square root of b^2 - 4ac/2a
I think the same method to factor but use the quadratic formula to get the roots will get the answer.
(I don't get you here)I also think the book meant to solve this problem instead.
I am very sure!2^x + 2^-x = 2
The book is wrong, I will be inclined to say that, what do you think should be the method that will eventually give you the correct answer?
2^0 - 2^ 0 =
1 - 1 = 0 I got zero
Yes that is exactly what's in the book, I have double check it.
The book has at it back zero [0] as its answer, please do it for me to see by using log, I am certain you can get by log application
Thanks!
Following the question;
= 2^2x - 1 = 2^(1+x)
= (2^x)^2 - 2^(1+x) = 0
Let 2^x = N
= N^2 - 2N - 1 = 0
Factorizing impossible!
What do you say?
If factorization is impossible should one use log? Or the method of completing the square?