# Logarithm

• Oct 4th 2010, 05:27 PM
Hellbent
Logarithm
Hi,

Are there alternative ways of solving this equation? Could logarithms be used?

$a^{60} = 2.044 \implies a = 2.044^{1/60} = 1.012$
• Oct 4th 2010, 05:31 PM
skeeter
Quote:

Originally Posted by Hellbent
Hi,

Are there alternative ways of solving this equation? Could logarithms be used?

$a^{60} = 2.044 \implies a = 2.044^{1/60} = 1.012$

logs could be used ... but the way you did it more direct.
• Oct 4th 2010, 05:38 PM
Hellbent
I need assistance with the logs part, please.
Tried using logs prior to posting question, but the answers are inconsistent.
• Oct 4th 2010, 05:51 PM
bigwave
Quote:

Originally Posted by Hellbent
Hi,

Are there alternative ways of solving this equation? Could logarithms be used?

$a^{60} = 2.044 \implies a = 2.044^{1/60} = 1.012$

$60\log{a} = \log{2.044}$

$\log{a} = \frac{log{2.044}}{60}
\RightArrow
= .00517
$

so $10^{(.00517)} = 1.01199$ or $1.012$

as mentioned your first method is better..
• Oct 4th 2010, 05:53 PM
skeeter
$a^{60} = k$ , where $k = 2.044$

$60\log{a} = \log{k}
$

$\displaystyle \log{a} = \frac{\log{k}}{60}$

$\displaystyle a = e^{\frac{\log{k}}{60}}$

$a \approx 1.012$

note that $\displaystyle a = e^{\frac{\log{k}}{60}} = \left(e^{\log{k}}\right)^{\frac{1}{60}} = k^{\frac{1}{60}} = 2.044^{\frac{1}{60}}$