Logarithm

• Oct 4th 2010, 04:27 PM
Hellbent
Logarithm
Hi,

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

$\displaystyle a^{60} = 2.044 \implies a = 2.044^{1/60} = 1.012$
• Oct 4th 2010, 04:31 PM
skeeter
Quote:

Originally Posted by Hellbent
Hi,

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

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

logs could be used ... but the way you did it more direct.
• Oct 4th 2010, 04:38 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, 04:51 PM
bigwave
Quote:

Originally Posted by Hellbent
Hi,

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

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

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

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

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

as mentioned your first method is better..
• Oct 4th 2010, 04:53 PM
skeeter
$\displaystyle a^{60} = k$ , where $\displaystyle k = 2.044$

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

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

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

$\displaystyle a \approx 1.012$

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