How can I solve this, when the indices are not equal.

Thanks! Sorry if this is a stupid question, but I'm studying to improve my math.


Let $\displaystyle(58)^a=(5.8)^b=10^c=k$

For real $a,b,c$ if $k=1, a=b=c=0$

Else $\displaystyle58=k^{1/a},5.8=k^{1/b},10=k^{1/c}$

$\displaystyle\implies k^{1/a}=58=5.8\cdot10=k^{1/b}\cdot k^{1/c}=k^{1/b+1/c}$

As $k\ne1,0$ we must have $\displaystyle\dfrac1a=\dfrac1b+\dfrac1c$

  • $\begingroup$ how could you take this k^1/a, as a=0, please elaborate it $\endgroup$ – Hemanta Paul Apr 9 '15 at 19:54
  • $\begingroup$ @HemantaPaul, Have you noticed "Else" $\endgroup$ – lab bhattacharjee Apr 10 '15 at 5:44

You have $58^a={(\frac {58}{10})}^b=10^c$ so let's start from the first two terms:

$58^a={(\frac {58}{10})}^b$

Taking log on both sides gives you:

$a=\log_{58}({(\frac {58}{10})}^b)$

Using log's properties leads you to:


Now we look at the second and third terms:

${(\frac {58}{10})}^b=10^c$

Same procedure gives you:

$c=b\log58-1$ (I omitted the base because it's $10$)

Now if you equal the first and the last terms you have:

$b={{\log_{58}10-1}\over {1-\log58}}$

Substituction leads you to:

$a={{\log_{58}10-1}\over {1-\log58}}-\log_{58}10$;$c={{\log_{58}10-1}\over {\log58 -1}}-1$

  • $\begingroup$ May you can simplify more the last results. $\endgroup$ – Renato Faraone Apr 9 '15 at 17:02

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