What are Some Solutions to Common Logarithm Problems?

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SUMMARY

This discussion addresses solutions to common logarithm problems, specifically focusing on equations involving base 10 and base 4 logarithms. The first equation, log10x - 2 = 0, yields a solution of x = 100. The second equation, ln(x + 5) = ln(x - 1) - ln(x + 1), results in no valid solution due to negative arguments in the logarithm. Lastly, the equation log4x - log4(x - 1) = 1/2 simplifies to x = 2, which is valid as x > 1.

PREREQUISITES
  • Understanding of logarithmic properties and identities
  • Familiarity with natural logarithms (ln) and common logarithms (log10)
  • Basic algebraic manipulation skills
  • Knowledge of logarithmic equations and their solutions
NEXT STEPS
  • Study the properties of logarithms, including change of base and product/quotient rules
  • Learn how to solve logarithmic equations involving multiple logarithmic terms
  • Explore the implications of negative arguments in logarithmic functions
  • Practice solving logarithmic equations with different bases, such as log2 and loge
USEFUL FOR

Students, educators, and anyone looking to enhance their understanding of logarithmic functions and solve logarithmic equations effectively.

mathdummy
log_10x-2=0 ... that's log base 10
(answer: 100)

ln(x+5)=ln(x-1)-ln(x+1)
(answer: no solution. but why?)

log_4x-log_4(x-1)=1/2
(answer: 2)
 
Last edited by a moderator:
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<br /> \ln(x+5)=\ln(x-1)-\ln(x+1)=\ln\left(\frac{x-1}{x+1}\right)<br />

So,

<br /> x+5=\frac{x-1}{x+1}<br />

Solve this for x, and you will see that both possible values generate a negative argument in the second logarithm, so there is no valid solution.
 
#1:


\log_{10} x - 2 = 0


10^{\log_{10}x} = 10^{2}


x = 10^{2} = 100

#3:


\log_{4}x - \log_{4}(x-1) = 0.5

\log_{4}\frac{x}{(x-1)} = 0.5

\frac{x}{(x-1)} = 2

x = 2x - 2

x = 2 (x &gt; 1)
 
Last edited:

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