What is the Correct Notation for Logarithmic Equations?

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Homework Help Overview

The discussion revolves around the correct notation and manipulation of logarithmic equations, specifically in the context of solving the equation (2^(x+1)) / (5^x) = 3.

Discussion Character

  • Conceptual clarification, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore various logarithmic manipulations and question the validity of their steps. There is a focus on the conditions under which logarithmic properties apply, particularly regarding subtraction and division of logs.

Discussion Status

Several participants have provided guidance on notation and suggested alternative approaches to the problem. There is an ongoing exploration of the implications of different logarithmic expressions and their interpretations, with no clear consensus reached yet.

Contextual Notes

Participants express confusion over notation and the proper application of logarithmic rules, indicating a need for clarity in mathematical expressions. There are references to specific examples that illustrate potential misunderstandings in logarithmic manipulation.

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Homework Statement



(2^(x+1)) / (5^x)) = 3


2.The attempt at a solution

(2^(x+1)) / (5^x)) = 3

((log 2 (x+1)) / (x log 5)) = log 3

(x log 2 + log 2) / (x log 5) = log 3

( log (2^ (x+1))) / (log (5^x)) = log 3

log (base: 5^x) (number: 2 ^ (x+1)) = log 3

(5^x) ^ (log 3) = 2 ^ (x+1)

(x log 5) ( log 3) = x log 2 + log 2

x ~ 9.09

I know that this is not the right answer but I don't understand where I went wrong.
 
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Hi ThatDude! :smile:

(try using the X2 button just above the Reply box :wink:)
ThatDude said:
(2^(x+1)) / (5^x)) = 3

((log 2 (x+1)) / (x log 5)) = log 3

((log 2 (x+1)) minus (x log 5)) = log 3 :smile:

btw, it would have been simpler to start by expressing the LHS as a multiple of (2/5)x :wink:
 
tiny-tim said:
Hi ThatDude! :smile:

(try using the X2 button just above the Reply box :wink:)


((log 2 (x+1)) minus (x log 5)) = log 3 :smile:

btw, it would have been simpler to start by expressing the LHS as a multiple of (2/5)x :wink:


Thank you for helping me out.

I thought that you could only subtract two logs when the number inside the log was being divided. In this case, isn't it the division of two logs, albeit with the same base?

Ex:
= log5 x/12

= log5x - log512

But from what I understand from your post, the following is also true:


log5 x/12 = log5x / log5 12
 
ThatDude said:
I thought that you could only subtract two logs when the number inside the log was being divided.

it is

you confused yourself by leaving out a step …

2x+1 / 5x = 3

log (2x+1 / 5x) = log 3

log (2x+1) - log (5x) = log 3 :wink:
 
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ThatDude said:
Thank you for helping me out.

I thought that you could only subtract two logs when the number inside the log was being divided. In this case, isn't it the division of two logs, albeit with the same base?

Ex:
= log5 x/12

= log5x - log512

But from what I understand from your post, the following is also true:


log5 x/12 = log5x / log5 12

You have some confusing expressions:

What does log5 x/12 mean?

Is it log (5x/12) or log_{5}(x/12)?
 
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ThatDude said:
Thank you for helping me out.

I thought that you could only subtract two logs when the number inside the log was being divided. In this case, isn't it the division of two logs, albeit with the same base?

Ex:
= log5 x/12

= log5x - log512

But from what I understand from your post, the following is also true:


log5 x/12 = log5x / log5 12

It is impossible to say whether you are correct, or not, because of the poor notation. If ##\log 5^{x/12}## means ##\log_5 (x/12)## then what you wrote is true. If it means ##\log \left( 5^{x/12}\right)## then what you wrote is false.
 
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tiny-tim said:
it is

you confused yourself by leaving out a step …

2x+1 / 5x = 3

log (2x+1 / 5x) = log 3

log (2x+1) - log (5x) = log 3 :wink:

Ah. I see. Thank you for your help.

Yes... Ray Vickson and SteamKing, it is log5(x/12) ; that is, the base is 5 and the argument is x divided by 12. I apologize for the poor notation.
 
ThatDude said:
Ah. I see. Thank you for your help.

Yes... Ray Vickson and SteamKing, it is log5(x/12) ; that is, the base is 5 and the argument is x divided by 12. I apologize for the poor notation.

In that case you should write something like log[5] x/12 or log_5 x/12 or log5 x/12; just about everybody would "get it" if you wrote it in any of these three ways (although it would not hurt to also include a brief verbal description, saying what the '5' means). It would be even better to include parentheses, like this: log[5] (x/12) or log_5 (x/12) or log5 (x/12).
 

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