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OpenStudy (anonymous):
\[\log_{2}(-13.4) \]
OpenStudy (anonymous):
i only know how to take the ln of negative numbers based on euler's formula
OpenStudy (anonymous):
because \[e^{\pi*i} = -1\]
OpenStudy (anonymous):
@jim_thompson5910 can you help me please?
jimthompson5910 (jim_thompson5910):
let's see how far we get using that equation
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jimthompson5910 (jim_thompson5910):
\[\Large e^{\pi*i} = -1\]
\[\Large 13.4*e^{\pi*i} = 13.4*(-1)\]
\[\Large 13.4*e^{\pi*i} = -13.4\]
see where I'm going with this?
OpenStudy (anonymous):
okay this makes sense. but how will i take \[\log_{2} \] and clear e?
jimthompson5910 (jim_thompson5910):
by using the change of base formula
\[\Large \log_{b}(x) = \frac{\ln(x)}{\ln(b)}\]
jimthompson5910 (jim_thompson5910):
the logs on the right side can be any base you want, in this case, they are base e
jimthompson5910 (jim_thompson5910):
just make sure that the logs on the right side have the same base
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OpenStudy (anonymous):
so then would i say:\[\log_{2}(-13.4) = \frac{ \ln (-13.4) }{ \ln (2) } \] \[\frac{ \ln (-1) * \ln(13.4) }{ \ln(2) }\] \[\frac{ i \pi * \ln(13.4) }{ \ln(2) }\]
and that would be my answer when i put the two natural logs in my calculator?
jimthompson5910 (jim_thompson5910):
close
jimthompson5910 (jim_thompson5910):
but you use the rule log(x*y) = log(x)+log(y)
jimthompson5910 (jim_thompson5910):
that means ln(-1*13.4) = ln(-1) + ln(13.4) = pi*i + ln(13.4)
OpenStudy (anonymous):
oh okay thank you very much!
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