Proof of logarithmic properties.

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SUMMARY

The discussion focuses on proving two logarithmic properties: 1) ln(b1/n) = (1/n)ln(b) for b > 0, with b set as an, and 2) ln(ar) = rln(a) for any rational r and a > 0. The key distinction is that the first property applies to rational exponents, while the second extends the proof to non-natural numbers. The user initially confuses the application of logarithmic properties for natural numbers with those for rational numbers, highlighting the need for a different approach when m is not a natural number.

PREREQUISITES
  • Understanding of natural and rational numbers
  • Familiarity with logarithmic properties, specifically ln(an) = nln(a)
  • Basic knowledge of exponential functions and their inverses
  • Concept of limits and continuity in calculus
NEXT STEPS
  • Study the proof of logarithmic properties for rational exponents
  • Learn about the continuity of logarithmic functions
  • Explore the relationship between logarithms and exponential functions
  • Investigate advanced logarithmic identities and their applications
USEFUL FOR

Students studying calculus or advanced algebra, particularly those focusing on logarithmic functions and their properties. This discussion is beneficial for anyone looking to deepen their understanding of logarithmic proofs and their applications in mathematics.

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Homework Statement



There are two log properties that I have to prove:
1) Explain why ln(b1/n)=(1/n)ln(b) for b>0, set b=an

2) Explain why ln(ar)=rln(a) for any r in Q and a>0, ie r is rational.

Homework Equations



ln(an)=nln(a)

The Attempt at a Solution



In a previous question I have already proved that ln(an)=nln(a), where n is a natural number. What I'm unsure about, is how is this any different? For 1), I'm not sure why you set b=an? Wouldn't you get ln((an)1/n) = ln(a)? I'm not sure how this helps me find the solution.

Similarly for 2), I'm unsure how it is any different to proving that ln(an)=nln(a).

Any help would be great!
 
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The obvious difference is that you proved a result m ln(a)=ln(a^m) when m is a natural number, and now they want you to do it when m is not a natural number. The proof that you used for this will not work when m is not a natural number (unless you did something really clever) because at some point you assume that it was, probably in order to write am as a*a*a...*a a multiplied together m times (which doesn't make sense if m=1/2 for example)

If b=an, what does ln(b) and ln(b1/n) become? Use it to do something cool
 
Office_Shredder said:
The obvious difference is that you proved a result m ln(a)=ln(a^m) when m is a natural number, and now they want you to do it when m is not a natural number. The proof that you used for this will not work when m is not a natural number (unless you did something really clever) because at some point you assume that it was, probably in order to write am as a*a*a...*a a multiplied together m times (which doesn't make sense if m=1/2 for example)

If b=an, what does ln(b) and ln(b1/n) become? Use it to do something cool

Thanks for your help. I worked out part 1, but I'm still unsure on part 2. For proving the law for a natural number I did:
Let y=ln(a), for a>0,
ey=a, and if we raise each side to the power n we get:
(ey)n=an
ln(ey*n)=ln(an)
y*n = ln(an), but y = ln(a)
n*ln(a) = ln(an)

So I'm still unsure how it's any different for any rational number?
Any help would be great :)
 
Use the definition of Q, namely that r =m/n, where m,n are in N.
 
Ah I get it now! Thanks :)
 

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