question about a logarithmic funtion using only variables

Question

what is the answer to this problem...

ab^cx+d +k = h

Answer 1

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.

Answer 2

(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

 

Answer 3

ab^cx+d+k=h

ab^cx+d=h - k

log_ab(h-k)=cx+d

log_ab(h-k)-d=cx

[log_ab(h-k)-d]/c=x