A line has equation In(y) = a In(x) + b where a = 0.796 and b = 3.538. The equation can be written in the form y = cx^d What is the value of c, to 2 decimal places? anyone can help wif this question?
Follow Math Help Forum on Facebook and Google+
I assume you mean $\displaystyle ln$ instead of $\displaystyle In$...as a hint, you may write: $\displaystyle \ln(y)=a\ln(x)+b=\ln(x^a)+\ln(e^b)=ln(e^bx^a)$
Hi ! ln(y) = a*ln(x) +b exp(ln(y)) = exp(a*ln(x) +b) exp(ln(y)) = exp(b)*exp(a*ln(x)) simplify exp(ln(y)) simplify exp(a*ln(x))
Originally Posted by MarkFL2 I assume you mean $\displaystyle ln$ instead of $\displaystyle In$...as a hint, you may write: $\displaystyle \ln(y)=a\ln(x)+b=\ln(x^a)+\ln(e^b)=ln(e^bx^a)$ if that is the case, $\displaystyle ln$ can be cancel off from both side which left with c = $\displaystyle e^b$?
Yes, now just use the given value for b and round to two decimal places as required.
thanks man!! then how about $\displaystyle ln$(y) = ax + b in the form of y=c$\displaystyle e$^dx
To rewrite the right side, use $\displaystyle u=\ln(e^u)$ then $\displaystyle \ln(v)+\ln(w)=\ln(vw)$.
Originally Posted by MarkFL2 To rewrite the right side, use $\displaystyle u=\ln(e^u)$ then $\displaystyle \ln(v)+\ln(w)=\ln(vw)$. can u explain in a simple way? i got messed up wif the v and w
Those were just meant as general variables... We are given: $\displaystyle \ln(y)=ax+b=\ln(e^{ax})+\ln(e^b)=\ln(e^be^{ax})$ Do you see how I applied the general rules I gave above? Can you finish?
Originally Posted by MarkFL2 Those were just meant as general variables... We are given: $\displaystyle \ln(y)=ax+b=\ln(e^{ax})+\ln(e^b)=\ln(e^be^{ax})$ and solve the remaining values Do you see how I applied the general rules I gave above? Can you finish? same with the above... Remove both side $\displaystyle \ln$
Yes, since logarithmic functions are monotonic (one-to-one), $\displaystyle \ln(u)=\ln(v)$ implies $\displaystyle u=v$.
View Tag Cloud