You are not logged in.
Wow!
I confronted a math tutor - but he claimed the answer 1/3 or -1/3 is correct - now I Could realise that your method is good. Please I want the tutor to see for himself your steps because he was insisting that the books' answer is correct
Thanks bobbym
Yes that's the problem - I copied it correctly.
When you plug in your answers does it give 1?
No, it doesn't give 1/3
Hi, Bobbym, I was called away, I am sorry - but I don't have my calculator with me now.
It is negative or positive 1/3 in the book as an answer. When I get to the house I will punch them in the calculator and see if it really gives negative or positive 1/3.
Yes I see now
It seems I am not getting you
Are we talking about log(x+1) - 2logx^2 = 1?
Yes.
The book tells that, if the base is not written then it in base 10. What do say? Again, are you saying it is a wrong answer?
[the book solved it]
log(x+1) - 2logx^2 = 1
log(x+1/x^2) = log10
Anti-log taking at both sides
x + 1 = 10x^2
9x^2 = 1
x^2 = 1/9
x = +1/3
This method is quite understandable to me, but to my suprise I can not use it to solve the one under discussion literally for many days
In fact, I am well familiar with my method - but unfortunately I can't arrive on 1/4 as the answer
Divide both sides by - 3
and the rest is easy.
Please - continue I am not getting the answer
Hi;
please - show the steps of my idea too.
I am very sure if I see your steps would help me know how to go about a similar problem.
Please, if you would not mind use my idea to solve, because I am rather familiar with that idea.
Thank you very much
Okay, I will let you see that problem yourself.
In one of the rules of logarithm it says, log(m/n) equal log m - log n, so I was expecting you would say logx - 4logx would be log(x/x^4). But couldn't understand why you subtracted logx from 4logx.
There was a similar problem in the book, and there was 1 at the right hand side of the equation, and in the course of manipulation it became [log10] and when antilog was taken it became 10. So that's the idea I tried using it. I don't know if you understand what I am putting across.
Hi; this is the problem I spoke of above, that used the one I understand;
log(x^2 + 1 ) - 2logx = 1
log(x^ + 1/ x^2) = log10
x^ + 1/x^2 = 10
x^2 + 1 = 10x^2
9x^2 = 1
x^2 = 1/9
x = +1/3
Please, see the method - this is the one I wanted to use for the one under discussion
Where does log8^2 come from?
That's according to the difinition of logarithm
- "the logarithm of any number b to the base a is the index or power to which the base must be raised to give the number" example ;
log b = c [taking 'a' as the base] = b = a^c
If so, then I was thinking the steps should be;
logx - 4logx = 2
log(x/x^4) = log8^2
taking antilog of both sides
x/x^4 = 8^2
x = 64x^4
What do you say to this?
I see now, but do you agree that a minus sign in logarithm means one has to divide?
logx - 4logx = 2
I'm now grasping it - how are you gonna proceed from here?
How I did it was;
log(x/x^4) = 8^2
taking antilog
x = 64x^4
[I was rather thinking that minus sign in logarithm means a division sign, what do you say to this?]
logx - 4logx = 2
wierd to me though
Like this?
x - x^4 = 8^2
Is 8 not going to raise the power 2 at the RHS?
Now before I proceed on the decimals, let me endeavour to put this across c'os I am getting zero as an answer to this and the book seems to have 1/4 ;
logx - 4logx = 2
The LHS is having 8 as their bases
Thank you!