6.09 Further applications of logarithms

Using our knowledge of the natural logarithm function and its inverse relationship with the exponential function, we can apply our newly acquired tools in solving equations and calculus to a wide range of applications. We can now find algebraic solutions for exponential problems where previously we required technology, as well as being able to apply calculus to general exponential functions. 

Keep the following rules in mind as we look at some examples.

Function	Derivative
$e^{f\left(x\right)}$ef(x)	$f\ '\left(x\right)e^{f\left(x\right)}$f ′(x)ef(x)
$\ln\left(f\left(x\right)\right)$ln(f(x))	$\frac{f\ '\left(x\right)}{f\left(x\right)}$f ′(x)f(x)​

 

Function $f\left(x\right)$f(x)	Integral $\int f\left(x\right)dx$∫f(x)dx
$e^{ax+b}$eax+b	$\frac{1}{a}e^{ax+b}+C$1a​eax+b+C
$\frac{1}{ax+b}$1ax+b​	$\frac{1}{a}\ln\left(ax+b\right)+C$1a​ln(ax+b)+C, for $ax+b>0$ax+b>0

 

Worked example

Example 1

The gradient of a curve at any point is given by $f\ '\left(x\right)=\frac{6}{x-3}$f ′(x)=6x−3​, and the curve passes through the point $\left(4,2\right)$(4,2) and is defined for $x>3$x>3.

Find an expression for the function $f(x)$f(x).

Think: Use integration to find the family of curves with the gradient function $f\ '\left(x\right)=\frac{6}{x-3}$f ′(x)=6x−3​ and then use the given point to solve for the constant of integration, $C$C.

Do:

$f\left(x\right)$f(x)	$=$=	$\int\frac{6}{x-3}dx$∫6x−3​dx
	$=$=	$6\int\frac{1}{x-3}dx$6∫1x−3​dx
	$=$=	$6\ln\left(x-3\right)+C$6ln(x−3)+C

Using the given point $\left(4,2\right)$(4,2), substitute into the function to solve for $C$C.

$2$2	$=$=	$6\ln\left(4-3\right)+C$6ln(4−3)+C
$2$2	$=$=	$6\ln1+C$6ln1+C
$\therefore$∴    $C$C	$=$=	$2$2

The function is given by $f\left(x\right)=6\ln\left(x-3\right)+2$f(x)=6ln(x−3)+2.

 

Practice questions

Question 1

Question 2