Expanding ln isn't working for me

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Homework Help Overview

The discussion revolves around the manipulation of logarithmic equations, specifically focusing on the expansion of natural logarithms and the application of exponentiation. The original poster presents an equation involving the natural logarithm and expresses confusion regarding the correct approach to solving it after expanding the logarithm.

Discussion Character

  • Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to solve an equation by expanding the logarithm but encounters discrepancies in the results. Some participants question the validity of the steps taken after expansion, particularly regarding the properties of logarithms and exponentiation.

Discussion Status

The discussion is ongoing, with some participants providing clarification on the properties of logarithms. There is an acknowledgment of the original poster's misunderstanding, but no consensus has been reached on the overall approach to the problem.

Contextual Notes

Participants are discussing the implications of expanding logarithmic expressions and the rules governing logarithmic identities, indicating a need for careful consideration of mathematical properties in the context of the problem.

iScience
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original equation: ln(0.2048x[itex]\frac{5}{2}[/itex])= -[itex]\frac{5}{2}[/itex]

when i just raise this to e i get..

0.2048T[itex]\frac{5}{2}[/itex]= -[itex]\frac{5}{2}[/itex]

and then i can solve for x and get the right answer. but when i expand the logs..

ln(0.2048) + [itex]\frac{5}{2}[/itex]ln(x)= -[itex]\frac{5}{2}[/itex]

0.2048+[itex]\frac{5}{2}[/itex]x=e-[itex]\frac{5}{2}[/itex]

then i try to solve for x i don't get the right answer. what am i doing wrong?
 
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you can't do that with a sum of ln's. Taking the exponent of both sides after expansion does not cancel out the ln in each term. You would have e^(lnA + lnB) and that does not equal A+B.
 
oops, lol thanks
 
no problem
 

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