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If we take it from the start. the n has to be squared when the square root at the R.H is taking away.
By the way what is mathematica I am curious to know.
the 'n' at the bottom carries a square why didnt you bring it?
Yes, correcto perfecto.
Oops. Bobym.
Please I made a little blunder!
2logy-log2x=log2(y-x).
Bobbym, I cant found them at the bottom of the open window. There are few that are found there.
But Bobbym I havent envisioned that bracket of the right hand side could cause any thing, but I think;
Log2(y-x) = log2y-log2x
I think is the same as, log(2y)-log(2x)
Please help if I am confused.
Thanks, Bobbym.
Then please let me how it should be
By the way I have developed the attitude of not putting brackets c'os the book I have doesn't put it around each example it gives. And that has become of me.
No.
the k squared plus the p squared are on one platform, all over "hg". That's all,
Errr, I dont normally put bracket around figures as you normally do.
But I suppose each of them are the same, I mean what yours above.
In #229
Make k the subject.
I had, k=(hg/n^2-p^2)
But the book has 1/n and a negative or positive sign after the "equl to sign", which is incomprehensible to me. Some help.
Please solve it and you will definitly come across what I mean.
it's log2(y-x)
I multiplied log2 with what is in the bracket, to have that. Why you say so.
Hi;
What comes after the k^?
"equal to sign" and after the sign, there is a "negative or positive sign" which confound me much.
The 'k' squared plus the 'p' squared are on one platform, which is over the 'hg'.
And all of them are in a square root sign.
Thanks.
I have solved it but this way:
2logy-log2x = log2(y-x).
log(y^2/2x) = log2y-log2x. this is where I think the divison must apply to the right hand ones.
(y/2x) =(2y-2x) Taking antilogs
(y^2/2x) = 2x(2y-2x)
y^2 = 4xy-4x^2
y^2-4xy+4x^2 = 0
y^2-2xy-2xy+4x^2=0
y(y-2x)-2x(y-2x)=0
(y-2x)(y-2x)=0
(y-2x)^2.
My question is why not log2y-2x is not made to be log(2y/2x), since it has that negative sign? In order to go by the law.
Please clear my doubt.
Thanks much.
Make k the subject.
1/n = (k^+p^2/hg)^1/2
I have k=(hg/n^2-p^2) as the answer. But the book solved beyond my final solution, and even had negative or positive sign in it.
please help me understand why its had that final answer, and as well how it came by the negative or positive sign.
Thanks in advance.
This is its final answer;
yes, yes, cube root instead.
But are they all right in that manner?
Okay, then is the following correct?
For instance.
y^1/3=a. would be a^
3
y^3 = a. square root of a would then be taken.
Is the above right?
I mean
1/3logp=1
logp^1/3=10
p =10^1/3
third root is required.
p=2.15.
I am not saying anyone is wrong but this suprise me, c'os I have applied what I am talking about for long.
See, I am taking aback as to why the answer is 1000.
Because I have learnt that any number raised to power either 1/3, 1/2, 1/4. is demanding the third the square and the fourth root of that number. I see it that if the answer is 1000 in the above problem then the question should have been:
3logp=1. so that it would be,
log p^3=log10. taking antilog.
p^3 = 10
p =10^3.
p=1000
Please, anyone correct me if I am wrong.
Thanks.
Thanks, Bobbym!
God bless.
1/3logp=1. Find the value of p.
Bobbym, how would one do this? I have tried and my answer was not the same as what the book has.
Bobym, please could you produce the methods here, if the log3^1 were to be log1/3. What would be the answer?
Thanks.
Anonynmstify, if I am getting you right I think youre right, but then when you work it down you would arrive on the division sign, watch:
2logy-log2x=log2(y-x).
logy^2 - log2x = log2y-2x. Now, there come a negative sign at the right equation.
I am thinking that, since there is a negative sign in both the right and the left equation, both sides must have that division sign respectively. But to my suprise only the left equation had the division sign. Still I am not clear.
Bobym let me know why +4.
I know the book is right, but I can not figure out as to why it applied the law of division on the left equation only. I could see the right equation also has a negative sign between it and I think should be
lo2y/log2x so that it would also have a division sign as the left equation has. Then from there one can work it down.
Bobbym, why do you think the division sign was denied on the right hand equation only?
Or is only left hand equations that must assume it? And not right hand equations?
But that was how it exists in the book and I tried to solve.
Think the following is how it came by its answder.
Logy^2-log2x=log2y-log2x
(y^2/2x)= 2x(2y-2x).
y^2= 4xy-4x^2
y^2= 4xy-4x^2
y^-4xy+4x^2=0 I think could be factorised,
My question is, as there became a negative sign between log2y-log2x I think should be log(2y/2x). As the law says logx-logy=(x/y).
But why only the law affected the left hand equation only I mean (y^2/2x) and not the right hand as well?
Please assist.
I
thanks, but please see the rest if they are right with their respective answers from the book, and my answers as well.