How do you find x in terms of A for this equation

2x+1=xlogA

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- May 21st 2006, 10:18 PMkingkaisai2Solving a logarithm equation
How do you find x in terms of A for this equation

2x+1=xlogA - May 22nd 2006, 01:03 AMticbolQuote:

Originally Posted by**kingkaisai2**

2x +1 = xLogA

Divide both sides by x,

(2x +1)/x = LogA

2x/x +1/x = LogA

2 +1/x = LogA

1/x = LogA -2

Since Log(10) = 1, then,

1/x = LogA -2Log(10)

1/x = LogA -Log(10^2)

1/x = LogA -Log(100)

1/x = Log(A/100)

1/x = Log(0.01A)

Take the reciprocals of both sides,

x = 1 / Log(0.01A) -------------------------answer. - May 22nd 2006, 02:58 AMCaptainBlackQuote:

Originally Posted by**ticbol**

10, which it may or may not do. But as this post is in the

College/University Level Math forum it would be safer to assume

natural logarithms.

It would be best not to assume anything about the base of logarithms intended,

it is probably clear to the original poster, but may not be obvious at second

hand.

RonL - May 22nd 2006, 04:12 AMticbolQuote:

Originally Posted by**CaptainBlack**

ln is natural log, or log to the base e.

I do not assume. That's what they are. College, University, whatever Level. - May 22nd 2006, 04:51 AMtopsquark
The previously posted solution seems unnecessarily complicated. Let's try:

-Dan

PS Ticbol, I learned logarithms the same way you did: log by itself should represent . Right or wrong many people in the field DO use log to represent ln. I think it's silly since if they mean ln they should just write it. However my personal preferences don't seem to have any impact on professional Math writing. (Yet. ;) ) - May 22nd 2006, 04:59 AMRich B.
Hi:

From 2x+1 = xlogA, xlog(A) - 2x = 1. Factoring gives x[log(A) - 2] = 1. Therefore x = 1/[log(A) - 2].

Regards,

Rich B. - May 22nd 2006, 05:05 AMticbolQuote:

Originally Posted by**topsquark**

My old Calculus textbook calls log as log to the base e. I just do not believe that book on that. I do not have to conform to everything that book, or any book, says. I always believe what I want to believe.

Believe me, I am stubborn. I let nobody or whatbody (:-)) dictates me what to believe. - May 22nd 2006, 05:07 AMtopsquarkQuote:

Originally Posted by**Rich B.**

-Dan - May 22nd 2006, 06:14 AMCaptainBlackQuote:

Originally Posted by**ticbol**

normalisation of the Fourier Transform. There are a number of conventions

in use in different fields, ignoring it won't make it go away.

This link is to the Wikipedia article which explains the ambiguous usage of log

RonL - May 22nd 2006, 11:51 AMticbolQuote:

Originally Posted by**CaptainBlack**

It is no big deal to me. I am used to it.

Just because those guys use log to mean ln to stay with the Math society does not mean I join them. No. To or for me, log is to the base 10 and ln is to the base e.

In my everyday calculator, log is to the base 10 and ln is to the base e. It is a Texas Instrument model.

I also have a TI-86, which I do not know how to use, which I am lazy enough to learn how to use, where log is to the base 10 and ln is to the base e.

Wikipedia is more believable or credible than a TI calculator to students of Math? Crazy comparison, I know, but that is the point. Mathematicians do not need to appear "higher" than us students of Math that they have to confuse us with log for ln.

I do not allow them Mathematicians to fool me, anyway. Nope.

Log is to the base 10. Ln is to the base e. Let them wikipedia authors and them Mathematicians cry.