1. Solving a logarithm equation

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

2x+1=xlogA

2. Originally Posted by kingkaisai2
How do you find x in terms of A for this equation

2x+1=xlogA
You have to isolate the x, of course.

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.

3. Originally Posted by ticbol
You have to isolate the x, of course.

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.
Except you have assumed that $\displaystyle \log$ denotes log to the base
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

4. Originally Posted by CaptainBlack
Except you have assumed that $\displaystyle \log$ denotes log to the base
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
log is log to the base ten
ln is natural log, or log to the base e.
I do not assume. That's what they are. College, University, whatever Level.

5. The previously posted solution seems unnecessarily complicated. Let's try:

$\displaystyle 2x+1 = x \, log A$

$\displaystyle 2x + x \, log A = -1$

$\displaystyle x(2 + log A) = -1$

$\displaystyle x = \frac{-1}{2 + log A}$

-Dan

PS Ticbol, I learned logarithms the same way you did: log by itself should represent $\displaystyle log_{10}$. 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. )

6. 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.

7. Originally Posted by topsquark
The previously posted solution seems unnecessarily complicated. Let's try:

$\displaystyle 2x+1 = x \, log A$

$\displaystyle 2x + x \, log A = -1$

$\displaystyle x(2 + log A) = -1$

$\displaystyle x = \frac{-1}{2 + log A}$

-Dan

PS Ticbol, I learned logarithms the same way you did: log by itself should represent $\displaystyle log_{10}$. 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. )
I do not know about "Professional Math writing", but I believe log is log to the base 10, and ln is log to the base e.

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.

8. Originally Posted by Rich 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.
It was too cold last night. My Math skills are frozen! (Apologies!)

-Dan

9. Originally Posted by ticbol
log is log to the base ten
ln is natural log, or log to the base e.
I do not assume. That's what they are. College, University, whatever Level.
We may not like it but the situation with logs is as bad as it is with the
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

10. Originally Posted by CaptainBlack
We may not like it but the situation with logs is as bad as it is with the
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
Well, let the wikipedia people say that. Like I said, my Calculus book, calls log as log to the base e also. I do not believe those. I do not ignore those either. I just look at the context of what is written and if I see that the log there is supposed to be ln, then I see that log as ln. Just so I can follow what is being said.
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.