what is the answer to this problem...
ab^{cx+d} +k = h
what is the answer to this problem...
ab^{cx+d} +k = h
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.
(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
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
Comments
This "answer" was not intended as an answer to the problem, but as a comment on Casey V's answer.