Logarithm

**Hellbent** · October 4th 2010, 04:27 PM

Hi,

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

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

**skeeter** · October 4th 2010, 04:31 PM

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.

**Hellbent** · October 4th 2010, 04:38 PM

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

**bigwave** · October 4th 2010, 04:51 PM

Originally Posted by Hellbent

Hi,

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

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

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

$\log{a} = \frac{log{2.044}}{60}<br /> \RightArrow<br /> = .00517<br />$

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

as mentioned your first method is better..

**skeeter** · October 4th 2010, 04:53 PM

$a^{60} = k$ , where $k = 2.044$

$60\log{a} = \log{k}<br />$

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

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

$a \approx 1.012$

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

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