Simplify the following:

bobbym · 2013-07-27 11:31:58

Hi;

I did that to get x all by itself on the left. What happened is I subtracted log(2) from both sides. That moved it over to the right side leaving a simpler expression on the left.

In case you are having some problems with balancing an equation check this out.

http://www.mathsisfun.com/algebra/add-s … lance.html

EbenezerSon · 2013-07-27 21:05:52

Thank you very much bobbym for the link.
But there was xlog3 also present at the left side with the log2. Why didn't you balance the eqution with the log3 instead? because it was also present at the left side.

I just want to why you didnt use the log3 instead.

Thanks.

bobbym · 2013-07-27 21:09:20

Hi;

If I were to move the x log(3) to the right side I would have to subtract x log(3) from both sides. Now I would have x on both sides. Remember I am trying to get x all byt itself and on one side.

EbenezerSon · 2013-07-27 21:43:27

Okay, but bobbym, the law logx-logy= logx/logy.
If it had been, logy-logy=0, is the zero?

If the logs have the same figures, it not right for one to divide?
But rather subtraction must take place?

EbenezerSon · 2013-07-27 22:04:02

You had your final answer by this:

x = log2/log2+log3
= 1.477

But I cant tell why the has, 0.39

EbenezerSon · 2013-07-27 22:16:07

Please, see how I solved the following:

2^x +2^-x=2.
bases are the same, therefore equate the exponents.
x-1-x =1
= x-x-1=1
=0-1 =1
=0 = 1+1
=0=2.

Is that right?
I think bases are the same, so doesnt need logarithm application.

Last edited by EbenezerSon (2013-07-27 22:35:16)

bobbym · 2013-07-27 23:08:11

Hi;

What answer did you get for x?

I would do it like this:

Multiple everything by 2^x to clear fractions out.

Factor the left side.

So

(2x-1) = 0 and (2x-1) =0

2x-1 = 0 then x = 0

x = 0 is the answer.

EbenezerSon · 2013-07-28 01:59:51

I had zero as the answer.

But as I am looking through the book, I chanced on the following, which I have solved but one of the answers was impossible to get. And the book had zero with that answer.

3^x + 3^-1=4.

My final answers are, x=1, and the impossible answer, 3^x=4.

bobbym · 2013-07-28 02:03:09

Hi;

I am afraid x = 1 is incorrect.

EbenezerSon · 2013-07-28 04:28:04

The book has x=0 or 1 as answer at its back.

EbenezerSon · 2013-07-28 04:32:22

How then would one solve this?

EbenezerSon · 2013-07-28 05:37:05

Okay, I admit, then see my thoughful manipulation:

3^x + 3^x-1= 4
multiplying 3 through.
3*3^x + 3^x*1/3 = 4* 3.
Let 3^x=m. therefore,
=3m + m = 12
= 4m - 12 =
0

factorising.
= 4(m-3) =0.
Is this correct?

bobbym · 2013-07-28 12:00:42

Hi;

What did you get for x?

EbenezerSon · 2013-07-29 07:55:32

The final result is my answer

EbenezerSon · 2013-07-29 07:59:43

I have considered it carefully, but see no way to solve than what I have done, please assist.

Thanks.

bobbym · 2013-07-29 08:18:27

Hi;

Factor out a 3^(x-1) from each term on the left.

Divide both sides by 4.

Take log of both sides, you get.

(x-1) log(3) = 0

Divide both sides by log(3)

x - 1 = 0

x = 1

See you later, I need to do some chores and shopping.

EbenezerSon · 2013-07-29 09:21:50

bobbym wrote:

Hi;
Factor out a 3^(x-1) from each term on the left.

I don't understand how you factored it out.

bobbym · 2013-07-29 15:27:17

Hi;

3* 3^(x-1) = 3^x

EbenezerSon · 2013-08-03 06:02:36

Please check, the following for me.

Log(2x+3)=log4.
I had 20 as the answer, the has 10.

If 2logy-log2x=2log(y-x), express y in terms of x. I had (y-x)(y-2x^2). The book has y^-4xy+4^2x=0.

log(10+9x)-log(11-x)=2. I had 20 as answer, the book has 10.

Last but not the least
log81/log3^1.
the book has -4 at its back as the answer.

Thanks.

Please, check the above if the answers are right. Me vs the book. -:)

Last edited by EbenezerSon (2013-08-03 07:20:08)

bobbym · 2013-08-03 06:06:26

Hi;

Log(2x+3)=log4.
I had 20 as the answer, the has 10.

You had what for x?

EbenezerSon · 2013-08-03 06:24:05

I had 20.
The book has 10 as the answer
Sorry I supposed to have checked the writings before posting.

Thanks.

bobbym · 2013-08-03 06:25:30

Both answers are wrong. The correct answer is 1.

EbenezerSon · 2013-08-03 06:29:19

how about if the log4 were to be log45. What would the answer be?

Thanks.

bobbym · 2013-08-03 06:34:50

You mean log(45) then x = 21.

EbenezerSon · 2013-08-03 06:53:42

I
thanks, but please see the rest if they are right with their respective answers from the book, and my answers as well.

Math Is Fun Forum

#176 2013-07-27 11:31:58

Re: Simplify the following:

#177 2013-07-27 21:05:52

Re: Simplify the following:

#178 2013-07-27 21:09:20

Re: Simplify the following:

#179 2013-07-27 21:43:27

Re: Simplify the following:

#180 2013-07-27 22:04:02

Re: Simplify the following:

#181 2013-07-27 22:16:07

Re: Simplify the following:

#182 2013-07-27 23:08:11

Re: Simplify the following:

#183 2013-07-28 01:59:51

Re: Simplify the following:

#184 2013-07-28 02:03:09

Re: Simplify the following:

#185 2013-07-28 04:28:04

Re: Simplify the following:

#186 2013-07-28 04:32:22

Re: Simplify the following:

#187 2013-07-28 05:37:05

Re: Simplify the following:

#188 2013-07-28 12:00:42

Re: Simplify the following:

#189 2013-07-29 07:55:32

Re: Simplify the following:

#190 2013-07-29 07:59:43

Re: Simplify the following:

#191 2013-07-29 08:18:27

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#192 2013-07-29 09:21:50

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#193 2013-07-29 15:27:17

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#194 2013-08-03 06:02:36

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#195 2013-08-03 06:06:26

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#196 2013-08-03 06:24:05

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#197 2013-08-03 06:25:30

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#198 2013-08-03 06:29:19

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#199 2013-08-03 06:34:50

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#200 2013-08-03 06:53:42

Re: Simplify the following:

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