Simplifying Logarithms: Troubleshooting Common Mistakes

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SUMMARY

The discussion centers on the simplification of logarithmic expressions, specifically the equation y=ln(2x). The user initially misapplies the properties of logarithms, leading to two incorrect equations. The correct simplification involves recognizing that e^{ln(2) + ln(x)} equals e^{ln(2)} * e^{ln(x)}, which simplifies to 2x, not 2 + x. This highlights the importance of accurately applying logarithmic identities in mathematical simplifications.

PREREQUISITES
  • Understanding of logarithmic properties, specifically the product rule.
  • Familiarity with natural logarithms (ln) and exponential functions (e).
  • Basic algebra skills for manipulating equations.
  • Knowledge of mathematical notation and symbols used in logarithmic expressions.
NEXT STEPS
  • Study the properties of logarithms, focusing on the product, quotient, and power rules.
  • Practice simplifying logarithmic expressions using various examples.
  • Learn about the relationship between logarithms and exponential functions.
  • Explore common mistakes in logarithmic simplifications and how to avoid them.
USEFUL FOR

Students, educators, and anyone looking to strengthen their understanding of logarithmic functions and their applications in mathematics.

Rectifier
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Hey there!
I am getting two completely different equations when I try to simplify one.
What am I doing wrong?

1.
[tex]y=ln(2x) \ \Leftrightarrow \ y=ln(2) + ln(x) \ \Leftrightarrow \ e^y=e^{ln(2)}+e^{ln(x)} \ \Leftrightarrow \ e^y=2+x[/tex]

2.
[tex]y=ln(2x) \ \Leftrightarrow \ e^y=e^{ln(2x)} \ \Leftrightarrow \ e^y=2x[/tex]

I am sorry if its something completely obvious. Its pretty late here so my brain doesn't function properly :)
 
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Rectifier said:
Hey there!
I am getting two completely different equations when I try to simplify one.
What am I doing wrong?

1.
[tex]y=ln(2x) \ \Leftrightarrow \ y=ln(2) + ln(x) \ \Leftrightarrow \ e^y=e^{ln(2)}+e^{ln(x)} \ \Leftrightarrow \ e^y=2+x[/tex]
It should be ##e^{ln(2) + ln(x)} = e^{ln(2)} \cdot e^{ln(x)} ## in your third step.
Rectifier said:
2.
[tex]y=ln(2x) \ \Leftrightarrow \ e^y=e^{ln(2x)} \ \Leftrightarrow \ e^y=2x[/tex]

I am sorry if its something completely obvious. Its pretty late here so my brain doesn't function properly :)
 
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Oh! Thank you Mark!
 

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