0:00 / 0:00

Logarithmic Functions

For an exponential ax,(a>0)a^x,(a>0) to have an inverse function, we needa1a ≠ 1. In this case, the inverse function is f(x)=logax,f(x)=\log _ax,which has domain x>0x>0, is defined by the following relationship
logax=y    ay=x\boxed{\log_{a} x=y \iff a^y=x}


f(x)f(x) has a vertical asymptote at
x=0

The Natural Logarithm has a base of ee: lnx=logex\ln x =\log_e x

PAGE BREAK

Properties of Logarithms

For x,y,cI ⁣R  and  x,y>0x, y, c\in\R \;\text{and} \;x,y>0
  1. log1=0\log1=0
  2. log(xy)=log(x)+log(y)\log(xy) = \log(x) + \log(y)
  3. log(xy)=log(x)log(y)\log\left( \frac{x}{y} \right) = \log(x) - \log(y)
  4. log(xc)=clog(x)\log(x^c) = c\log(x)
  5. logaax=x\log_aa^x = x
  6. alogax=xa^{\log_ax} = x
  7. logcx=lnxlnc\displaystyle \log_c x=\frac{\ln x}{\ln c}
0:00 / 0:00

Example: Evaluating Logarithms

Evaluate log2 116log215+log230\log_2\ \frac{1}{16}-\log_215+\log_230

Let x=log2116log215+log230x=\log_2\frac{1}{16}-\log_215+\log_230
Using log rules, we can combine the right hand side
x=log2(116÷15×30)x=\log_2\left(\frac{1}{16}\div15\times30\right)
x=log2(116×15×30)x=\log_2\left(\frac{1}{16\times15}\times30\right)
x=log2(18)x=\log_2\left(\frac{1}{8}\right)

Using the relationship between log and exponent, we can rewrite this
2x=182^x=\frac{1}{8}
2x=232^x=2^{-3}
So, x=3x=-3.
Therefore, the expression evaluates to 3-3
0:00 / 0:00

Example: Inverse of Logarithms

Find the inverse of f(x)=ln(x4)f\left(x\right)=\ln\left(x-4\right) and state the domain and range of both f(x)f(x) and f1(x)f^{-1}(x).

Steps:
1. Write y=f(x)y=f\left(x\right)
y=ln(x4)y=\ln\left(x-4\right)

2. Solve for xx
ey=eln(x4)e^y=e^{\ln\left(x-4\right)}
ey=x4e^y=x-4
ey+4=xe^y+4=x
x=ey+4x=e^y+4

3. Swap xx and yy
y=ex+4y=e^x+4

4. Replace yy with f1(x)f^{-1}\left(x\right), this is the inverse function
Therefore, f1(x)=ex+4f^{-1}\left(x\right)=e^x+4

Domain and range:
𝑓 has domain (4,)\left(4,\infty\right) and range (,)\left(-\infty,\infty\right)
f1f^{-1} has domain (,)\left(-\infty,\infty\right) and range (4,)\left(4,\infty\right)
Solve for xx in the equation 2lnx=ln4+ln(x1)2\ln x=\ln4+\ln\left(x-1\right).

Extra Practice