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Solve for t using logarithms with base a
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\[\frac{ A-D }{ B }=a^Ct\]
oh snap the t should also be an exponent
\[\frac{ A-D }{ B }=a ^{ct}\]
\[\frac{ A-D }{ B }=a^{Ct}\] taking the log to base \(a\) on both sides \[\log_a\frac{ A-D }{ B }=\log_aa^{Ct}\] now use \(\boxed{\log_bb^n = n\log_bb = n}\)
you take the log of both sides?
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so the right side would be ct?
yepp
Now, to solve for \(t\), divide by \(C\)
\[\frac{ 1 }{ c }\log_{a} \frac{ A-D }{ B }=t\]
good work
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:D thank you so much
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