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# question about a logarithmic funtion using only variables

what is the answer to this problem...

abcx+d +k = h

### 3 Answers by Expert Tutors

David U. | Dave the Math TutorDave the Math Tutor
4.9 4.9 (238 lesson ratings) (238)
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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.

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

George C. | Humboldt State and Georgetown graduateHumboldt State and Georgetown graduate
5.0 5.0 (2 lesson ratings) (2)
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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

Casey V. | Basic math to trigonometry tutoring! Basic math to trigonometry tutoring!
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abcx+d+k=h

abcx+d=h - k

logab(h-k)=cx+d

logab(h-k)-d=cx

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