1. ## Logarithm

Hi,

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

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

2. 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.

3. I need assistance with the logs part, please.
Tried using logs prior to posting question, but the answers are inconsistent.

4. 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..

5. $\displaystyle a^{60} = k$ , 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 $\displaystyle \displaystyle a = e^{\frac{\log{k}}{60}} = \left(e^{\log{k}}\right)^{\frac{1}{60}} = k^{\frac{1}{60}} = 2.044^{\frac{1}{60}}$