Search 74,393 tutors
FIND TUTORS
Ask a question
0 0

question about a logarithmic funtion using only variables

Tutors, please sign in to answer this question.

3 Answers

Sorry, but this answer is wrong.  The exponent (cx+d) applies only to the b; the a is not part of the base.  So taking the base (ab) logarithm is the wrong move.  George C's answer, complicated as it is, is correct.

Comments

This "answer" was not intended as an answer to the problem, but as a comment on Casey V's answer.

(h - k)/a  =  b^(cx + d)

ln ((h - k)/a) =  (cx + d) ln b

(ln ((h - k)/a)/ln b) = cx + d

 (ln ((h - k)/a)/ln b)  -  d  =  cx

 ((ln ((h - k)/a)/ln b) - d)/c  =  x 

(((ln ((h - k)/a)))/ln b) - d)/c  =  x