Solve Logarithm Equation: 3x + 3x+1 = 4x

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Homework Help Overview

The problem involves solving the logarithmic equation 3x + 3x+1 = 4x. Participants are exploring the validity of applying logarithmic properties to manipulate the equation.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the incorrect application of the logarithmic property log(a + b) = log(a) + log(b) and question how to properly approach solving the equation. Some suggest factoring and using properties of exponents.

Discussion Status

The discussion includes multiple interpretations of the logarithmic properties and their application. Some participants have offered hints and suggestions for factoring, while others emphasize the need for correct assumptions regarding the logarithmic functions.

Contextual Notes

There is an ongoing debate about the assumptions related to the terms in the equation, particularly regarding the conditions under which certain logarithmic properties hold true.

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Homework Statement


3x + 3x+1 = 4x

Can I do this ?

Log 3x + Log 3x+1 = Log4x

thanks
 
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No, you can't. The law

\log(a+b)=\log(a)+\log(b)

is not valid.
 
scientifico said:

Homework Statement


3x + 3x+1 = 4x

Can I do this ?

Log 3x + Log 3x+1 = Log4x

thanks

Absolutely not. log(3^x)+log(3^(x+1))=log(3^x*3^(x+1)), not log(3^x+3^(x+1)).
 
How can I solve my equation ?
 
hi scientifico! :smile:
scientifico said:
Can I do this ?

Log 3x + Log 3x+1 = Log4x

no!

log(a + b) is not loga + logb !​

hint: factorise 3x + 3x+1 :wink:
 
scientifico said:
How can I solve my equation ?

3^(x+1)=3*3^x. Try using that.
 
Thanks I solved.
 
3(1-3^x) < 5^x(1-3^x)

must I impose 1-3^x > 0 and 1-3^x < 0 ?
 
scientifico said:
3(1-3^x) < 5^x(1-3^x)

must I impose 1-3^x > 0 and 1-3^x < 0 ?

Please post new questions in a separate thread.
 

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