Simplify the following:

bobbym · 2014-10-04 06:37:49

Wait, yours is getting the right answer too. My apology.

EbenezerSon · 2014-10-04 06:54:04

I am very sure if I see your steps would help me know how to go about a similar problem.

bobbym · 2014-10-04 06:57:57

Divide both sides by - 3

and the rest is easy.

EbenezerSon · 2014-10-04 07:01:08

Hi;
please - show the steps of my idea too.

bobbym · 2014-10-04 07:10:18

I am sorry but that I can not do since I do not understand it.

EbenezerSon · 2014-10-14 09:22:55

bobbym wrote:

Divide both sides by - 3
and the rest is easy.

Please - continue I am not getting the answer

bobbym · 2014-10-14 09:26:05

Raise both sides to the 8^ that will make the

disappear.

x = 8^(-2/3)

EbenezerSon · 2014-10-16 19:07:15

In fact, I am well familiar with my method - but unfortunately I can't arrive on 1/4 as the answer

bobbym · 2014-10-16 20:50:15

If it can not get the correct answer then either you are not using it right or the the method is not that good.

EbenezerSon · 2014-10-17 08:12:51

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

bobbym · 2014-10-17 08:55:42

Why do you think those logs are to the base 10?

If they are it is going to lead to a cubic in which 1 / 3 is not an answer.

EbenezerSon · 2014-10-19 18:06:08

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]

Last edited by EbenezerSon (2014-10-19 18:08:12)

bobbym · 2014-10-19 22:13:26

Are we talking about log(x+1) - 2logx^2 = 1?

EbenezerSon · 2014-10-20 00:08:09

bobbym wrote:

Are we talking about log(x+1) - 2logx^2 = 1?

Yes.

bobbym · 2014-10-20 00:12:37

If x = 1 / 3 is a solution when you plug in the equation should now be true.

EbenezerSon · 2014-10-20 00:18:18

It seems I am not getting you

bobbym · 2014-10-20 00:24:27

Okay, let us do something simpler and then we will come back to this.

When we say we solved an equation it means we have found the x value that makes the equation true.

Supposing we have this equation:

x + 5 = 10

if we now said that x = 6 is a solution when we plug in for x we get

6 + 5 does not equal 10 so we have not solved the equation. x = 6 is not a solution.

If we say x = 5 is a solution when we plug in we find

5 + 5 = 10 this is true! So, x = 5 is a solution to the equation x + 5 = 10.

Do you follow?

EbenezerSon · 2014-10-20 00:27:23

Yes I see now

bobbym · 2014-10-20 00:35:05

Now, the same thing holds for more complicated equations. If x = 1 / 3 was a solution to log(x+1) - 2logx^2 = 1 then when we plug in or replace x by 1 / 3 in it the statement should be true. It does not matter what method we use to ge this value of x, the plug in rule must always apply.

Now when you plug in x = 1 / 3 into log(x+1) - 2logx^2 = 1 is that true?

EbenezerSon · 2014-10-20 04:55:26

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.

Last edited by EbenezerSon (2014-10-20 04:56:30)

bobbym · 2014-10-20 05:01:19

See if log(x+1) - 2logx^2 equals 1.

EbenezerSon · 2014-10-21 03:42:25

No, it doesn't give 1/3

bobbym · 2014-10-21 03:49:09

So we know that x = 1 /3 or x = - 1 / 3 is not a solution.

When I solve for that I get a very complicated expression. This is not likely for a textbook.

Did you copy the problem correctly?

EbenezerSon · 2014-10-21 03:58:49

Yes that's the problem - I copied it correctly.

When you plug in your answers does it give 1?

bobbym · 2014-10-21 04:04:26

Yes, it does.

Math Is Fun Forum

#651 2014-10-04 06:37:49

Re: Simplify the following:

#652 2014-10-04 06:54:04

Re: Simplify the following:

#653 2014-10-04 06:57:57

Re: Simplify the following:

#654 2014-10-04 07:01:08

Re: Simplify the following:

#655 2014-10-04 07:10:18

Re: Simplify the following:

#656 2014-10-14 09:22:55

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#657 2014-10-14 09:26:05

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#658 2014-10-16 19:07:15

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