Express $log_3(a^2 + \sqrt{b})$ in terms of m and k where $m = log_{3}a$

$k = log_{3}b$

Given this information I made $a = 3^m$

$b = 3^k$

Therefore = $log_{3} ((3^m)^2 + (3^k))^{\frac{1}{2}}$

= $log_{3} (3^{2m} + 3^{\frac{k}{2}})$

I don't know if I'm done or there is still more things I can simplify. Can anyone help please, thanks

  • $\begingroup$ Is 3 in base ??? $\endgroup$ – Archis Welankar Dec 10 '15 at 15:58
  • $\begingroup$ Yes it's in the base $\endgroup$ – Question asker Dec 10 '15 at 16:01
  • $\begingroup$ Any ideas you have on simplification $\endgroup$ – Archis Welankar Dec 10 '15 at 16:05
  • $\begingroup$ Maybe the log 3 and the two 3's cancel out leaving only the exponents $\endgroup$ – Question asker Dec 10 '15 at 16:06
  • $\begingroup$ So its 2m + k/2 $\endgroup$ – Question asker Dec 10 '15 at 16:07

Since there is no expansion for $\log(a+b)$, hence, you could stop there. If you need to get an approximation(sometimes in computer science), you could use $\log(a+b)=\log a+\log(1+b/a)$, then you have following: $$2m+\log_3(1+3^{k/2-2m}).$$

  • $\begingroup$ Makes sense, but doesn't express in terms of k and m mean that k and m should be the only terms in an expression? Maybe I'm just wrong $\endgroup$ – Question asker Dec 10 '15 at 16:10
  • $\begingroup$ @Questionasker: that's what they are here ! $\endgroup$ – Yves Daoust Dec 10 '15 at 16:12
  • $\begingroup$ @Questionasker: $\log$ is just an operator, not a variable etc, hence, it's enough to express it using $\log$. If you have more conditions on $k$ and $m$, then maybe you could simplify the expression without $\log$. $\endgroup$ – user297600 Dec 10 '15 at 16:13

Alternatively, $\log_3(9^m+\sqrt3^k)$, but this makes little difference.

You can't decompose a sum inside a logarithm.


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