0
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

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 $\displaystyle\dfrac1a=\dfrac1b+\dfrac1c$

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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:

$a=b-\log_{58}10$

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:

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

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.