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

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The discussion centers on solving the logarithmic equation 3^x + 3^(x+1) = 4^x. Participants clarify that the logarithmic property log(a + b) ≠ log(a) + log(b) is incorrectly applied in the initial approach. Instead, they suggest factoring the left side of the equation and using properties of exponents. A solution was found involving inequalities, leading to further questions about imposing conditions on the factors. The conversation emphasizes the importance of correctly applying logarithmic and exponential rules in solving equations.
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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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