what is the answer to this problem...

ab^{cx+d} +k = h

what is the answer to this problem...

ab^{cx+d} +k = h

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(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.