Logarithms: Driving in fog ...solve the equation for I(d)

[wind]

Hi, can someone check over my work? Thanks

Driving in fog at night greatly reduces the intensity of light from an approaching car. The relationship between the distance,d, in meters, that your car is from the approaching car and the intensity of light, I(d), in lumens (1m), at distance d, is giving by

\(\displaystyle \L\\\ d=-166.67 log\) \(\displaystyle \L\\\frac{I(d)}{125}\)

solve the equation for I(d)

\(\displaystyle \L\\\ d=log\) \(\displaystyle \L\\\frac{I(d)}{125}^{-166.67}\)
...
\(\displaystyle \L\\\ 10^{d}=\) \(\displaystyle \L\\\frac{I(d)}{125}^{-166.67}\)
...
\(\displaystyle \L\\\ 10^{d}*125=\) \(\displaystyle \L\\\ I(d)^{-166.67}\)
...
\(\displaystyle \L\\\ log 10^{d}*125=\) \(\displaystyle \L\\\log I(d)^{-166.67}\)
...
\(\displaystyle \L\\\ dlog 10*125=\) \(\displaystyle \L\\\ log I(d)^{-166.67}\)
...
\(\displaystyle \L\\\ dlog 1250=\) \(\displaystyle \L\\\ -166.67log I(d)\)
...
\(\displaystyle \L\\\frac{d log 1250}{ -166.67 log}=\) \(\displaystyle \L\\\ I(d)\)

 

[Unco]

Re: Logarithms: Driving in fog ...solve the equation for I(d

Hi, can someone check over my work? Thanks

Driving in fog at night greatly reduces the intensity of light from an approaching car. The relationship between the distance,d, in meters, that your car is from the approaching car and the intensity of light, I(d), in lumens (1m), at distance d, is giving by

\(\displaystyle \L\\\ d=-166.67 log\) \(\displaystyle \L\\\frac{I(d)}{125}\)

solve the equation for I(d)

\(\displaystyle \L\\\ d=log\) \(\displaystyle \L\\\frac{I(d)}{125}^{-166.67}\)
No; that would be \(\displaystyle \mbox{\(\frac{I(d)}{125}\)^{-166.67}\)

...
\(\displaystyle \L\\\ 10^{d}=\) \(\displaystyle \L\\\frac{I(d)}{125}^{-166.67}\)
...
\(\displaystyle \L\\\ 10^{d}*125=\) \(\displaystyle \L\\\ I(d)^{-166.67}\)
...
\(\displaystyle \L\\\ log 10^{d}*125=\) \(\displaystyle \L\\\log I(d)^{-166.67}\)
...
\(\displaystyle \L\\\ dlog 10*125=\) \(\displaystyle \L\\\ log I(d)^{-166.67}\)

No; \(\displaystyle \mbox{ \log{(x^py)} = p\log{x} + \log{y}\)

\(\displaystyle \L\\\ dlog 1250=\) \(\displaystyle \L\\\ -166.67log I(d)\)
...
\(\displaystyle \L\\\frac{d log 1250}{ -166.67 log}=\) \(\displaystyle \L\\\ I(d)\)

You need to become more familiar with the log rules, Wind. Moreover, just think about isolating I(d), rather than just using those rules willy nilly...

\(\displaystyle \mbox{d = -166.67\ln{\(\frac{I(d)}{125}\)} \Right \ln{\(\frac{I(d)}{125}\)} = \frac{d}{-166.67}\)

Then take exponentials (assuming the log is base 10):

\(\displaystyle \mbox{\frac{I(d)}{125} = 10^{\(\frac{d}{-166.67}\)}}\), etc.