How do you solve for t #K= H - Ca^t#?

1 Answer
mason m
Jan 28, 2016

#t=log_a((H-K)/C)#

Explanation:

First, isolate the term with #t#.

#K-H=-Ca^t#

Divide by #-C#.

#(H-K)/C=a^t#

To undo an exponential function like #a^t# in order to solve for #t#, we will have to use a logarithm. Logarithms and exponential functions are inverses.

#log_a((H-K)/C)=log_a(a^t)#

#t=log_a((H-K)/C)#

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