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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

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David U. | Dave the Math TutorDave the Math Tutor
4.9 4.9 (227 lesson ratings) (227)
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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.

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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!
0

abcx+d+k=h

abcx+d=h - k

logab(h-k)=cx+d

logab(h-k)-d=cx

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