1. ## Help in Calculus!!

1. 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?

2. ## Re: Help in Calculus!!

I assume you mean $ln$ instead of $In$...as a hint, you may write:

$\ln(y)=a\ln(x)+b=\ln(x^a)+\ln(e^b)=ln(e^bx^a)$

3. ## Re: Help in Calculus!!

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))

4. ## Re: Help in Calculus!!

Originally Posted by MarkFL2
I assume you mean $ln$ instead of $In$...as a hint, you may write:

$\ln(y)=a\ln(x)+b=\ln(x^a)+\ln(e^b)=ln(e^bx^a)$
if that is the case, $ln$ can be cancel off from both side which left with c = $e^b$?

5. ## Re: Help in Calculus!!

Yes, now just use the given value for b and round to two decimal places as required.

6. ## Re: Help in Calculus!!

thanks man!! then how about $ln$(y) = ax + b in the form of y=c $e$^dx

7. ## Re: Help in Calculus!!

To rewrite the right side, use $u=\ln(e^u)$ then $\ln(v)+\ln(w)=\ln(vw)$.

8. ## Re: Help in Calculus!!

Originally Posted by MarkFL2
To rewrite the right side, use $u=\ln(e^u)$ then $\ln(v)+\ln(w)=\ln(vw)$.
can u explain in a simple way? i got messed up wif the v and w

9. ## Re: Help in Calculus!!

Those were just meant as general variables...

We are given:

$\ln(y)=ax+b=\ln(e^{ax})+\ln(e^b)=\ln(e^be^{ax})$

Do you see how I applied the general rules I gave above?

Can you finish?

10. ## Re: Help in Calculus!!

Originally Posted by MarkFL2
Those were just meant as general variables...

We are given:

$\ln(y)=ax+b=\ln(e^{ax})+\ln(e^b)=\ln(e^be^{ax})$ 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 $\ln$

11. ## Re: Help in Calculus!!

Yes, since logarithmic functions are monotonic (one-to-one), $\ln(u)=\ln(v)$ implies $u=v$.