PROBLEM MA127 CRUX MATHEMATICORUM V47 No 6

 Pag 275 - En

Pag 276 - Fr

ANSWER CiP 

$\log_5 12=\frac{2\cdot a+b}{1-a}$

     Solution CiP

          We have $\log_5 12 =\log_5(2^2 \cdot 3)=\log_5 2^2+\log_5 3$, so

$\log_5 12 =2\cdot \log_5 2 + \log_5 3. \tag{1}$

          First, $\frac{1}{a}=\log_2 10=\log_2 (2\cdot 5)=\log_2 2 +\log_2 5=1+\frac{1}{\log_5 2}$ so

$\frac{1}{\log_5 2}=\frac {1}{a}-1$, whence it results

$\log_5 2 =\frac {a}{1-a}. \tag{2}$

           Secondly, $\frac{1}{b}=\log_3 10 =\log_3 2 +\log_3 5=\frac{\log_{10} 2}{\log_{10} 3}+\frac{1}{\log_5 3}$, so $\frac{1}{\log_5 3}=\frac {1}{b}-\frac {a}{b}$, whence it results

$\log_5 3=\frac{b}{1-a}. \tag{3}$

          Replacing formulas (2) and (3) in formula (1) we get the answer.

$\blacksquare$

          

REMARK CiP

          More briefly $\log_5 12=\frac{\log_{10} 12}{\log_{10} 5}=\frac{\log_{10} 2^2 \cdot 3}{\log_{10} \frac{10}{2}}=\frac{\log_{10} 2^2 +\log_{10} 3}{\log_{10} 10-\log_{10} 2}=\frac{2\cdot \log_{10} 2 +\log_{10} 3}{1-\log_{10} 2}$ and we get the same answer.

$\blacksquare \blacksquare$

 

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Tracking Number: 12831
Received:        Thu Jul 15 07:10:34 2021

From:  Petre Ciobanu
       Scoala Gimnaziala "Samuil Micu" SADU
       Sibiu, Romania
Email: ptr.ciobanu@gmail.com

Type:  Solve a MathemAttic Problem
       (problem MA127)

Files:
  LATEX_source_of_MA127.doc
  PROBLEM_MA127_CRUX_v47_n6.pdf


Comments:

Crux Mathematicorum <crux@cms.math.ca>	10:12 (acum 22 de minute)
către eu