# Logs and Exponents

• Jan 18th 2009, 11:22 AM
Twenty80
Logs and Exponents
I'm a little stumped here, as we have a constant to the power of an exponent.. How would I simplify this:

$\displaystyle 6^{\log_{2}n}$

--------------------------------------------------------------------------

I would also appreciate it if someone could tell me if the answer I came up with for the following is correct:
Simplify the following:
$\displaystyle \log_{10}(10(\log_{10}n^3)100^{2n})$

$\displaystyle \log_{10}(30\log_{10}n) + 4n$

I feel my answer isn't complete as I'm not too sure how to further simplify what's left of the addition operator.
• Jan 18th 2009, 03:22 PM
vincisonfire
For the first one you have
$\displaystyle 6^{log_2 n} = (2\cdot 3)^{log_2 n} = 2^{log_2 n} 3^{log_2 n} = n 3^{log_2 n}$
For the second
$\displaystyle \log_{10}(30\log_{10}n) + 4n = \log_{10}^2(n) + \log_{10}(3) + \log_{10}(10) + 4n = \log_{10}^2(n) + \log_{10}(3) + 1 + 4n$
• Jan 18th 2009, 03:32 PM
Mush
Quote:

Originally Posted by vincisonfire
For the first one you have
$\displaystyle 6^{log_2 n} = (2\cdot 3)^{log_2 n} = 2^{log_2 n} 3^{log_2 n} = n 3^{log_2 n}$
For the second
$\displaystyle \log_{10}(30\log_{10}n) + 4n = \log_{10}^2(n) + \log_{10}(3) + \log_{10}(10) + 4n = \log_{10}^2(n) + \log_{10}(3) + 1 + 4n$

Alternatively:

$\displaystyle \log_{2}{n} = \log_{6}n \times \log_{2}6$

Hence:

$\displaystyle 6^{\log_{6}n \times \log_{2}6}$

$\displaystyle (6^{\log_{6}n})^{\log_{2}6}$

$\displaystyle (n)^{\log_{2}6}$

$\displaystyle (n)^{\log_{2}(2\times3)}$

$\displaystyle (n)^{\log_{2}2 + \log_{2}3}$

$\displaystyle (n)^{1+ \log_{2}3}$
• Jan 18th 2009, 03:35 PM
vincisonfire
Quote:

Originally Posted by Mush
$\displaystyle (n)^{1+ \log_{2}3}$

Certainly a better answer.