make x the subject y=5(1-e^(-x)) first i expand it so y=5-5e^(-x) loge(y)=loge(5)+loge(e5x) loge(y)-loge(5)=5x x=loge(y)-loge(5)/5 ??
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Originally Posted by smmmc make x the subject y=5(1-e^(-x)) first i expand it so y=5-5e^(-x) loge(y)=loge(5)+loge(e5x) loge(y)-loge(5)=5x x=loge(y)-loge(5)/5 ?? (Anything from this step onwards is just to get the answer to whatever form you like) .
the answer says loge(5/5-y) ?
Dear smmmc, Of course, Therefore what Prove it had shown is the answer.
Originally Posted by smmmc the answer says loge(5/5-y) ? Which is one of the forms of the answer given by Prove It. And you have been told in another post what ln means. So what's your problem here?
Originally Posted by smmmc make x the subject y=5(1-e^(-x)) first i expand it so y=5-5e^(-x) loge(y)=loge(5)+loge(e5x) This is wrong. log(a+ b) is not equal to log(a)+ log(b) As others have said, the simplest thing to do is write y- 5= -5e^{-x+} and take the logarithm of that. loge(y)-loge(5)=5x x=loge(y)-loge(5)/5 ??
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