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Logarithmic Functions

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


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

The Natural Logarithm has a base of eee: ln⁡x=log⁡ex\ln x =\log_e xlnx=loge​x

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Properties of Logarithms
For x,y,c∈I ⁣R  and  x,y>0x, y, c\in\R \;\text{and} \;x,y>0x,y,c∈IRandx,y>0  
log⁡1=0\log1=0log1=0
log⁡(xy)=log⁡(x)+log⁡(y)\log(xy) = \log(x) + \log(y)log(xy)=log(x)+log(y)
log⁡(xy)=log⁡(x)−log⁡(y)\log\left( \frac{x}{y} \right) = \log(x) - \log(y)log(yx​)=log(x)−log(y)
log⁡(xc)=clog⁡(x)\log(x^c) = c\log(x)log(xc)=clog(x)
log⁡aax=x\log_aa^x = xloga​ax=x
alog⁡ax=xa^{\log_ax} = xaloga​x=x
log⁡cx=ln⁡xln⁡c\displaystyle \log_c x=\frac{\ln x}{\ln c}logc​x=lnclnx​

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Example: Evaluating Logarithms 
Evaluate log⁡2 116−log⁡215+log⁡230\log_2\ \frac{1}{16}-\log_215+\log_230log2​ 161​−log2​15+log2​30

Let x=log⁡2116−log⁡215+log⁡230x=\log_2\frac{1}{16}-\log_215+\log_230x=log2​161​−log2​15+log2​30
Using log rules, we can combine the right hand side
x=log⁡2(116÷15×30)x=\log_2\left(\frac{1}{16}\div15\times30\right)x=log2​(161​÷15×30)
x=log⁡2(116×15×30)x=\log_2\left(\frac{1}{16\times15}\times30\right)x=log2​(16×151​×30)
x=log⁡2(18)x=\log_2\left(\frac{1}{8}\right)x=log2​(81​)

Using the relationship between log and exponent, we can rewrite this
2x=182^x=\frac{1}{8}2x=81​
2x=2−32^x=2^{-3}2x=2−3
So, x=−3x=-3x=−3.
Therefore, the expression evaluates to −3-3−3

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Example: Inverse of Logarithms
Find the inverse of f(x)=ln⁡(x−4)f\left(x\right)=\ln\left(x-4\right)f(x)=ln(x−4) and state the domain and range of both f(x)f(x)f(x) and f−1(x)f^{-1}(x)f−1(x).

Steps:
1. Write y=f(x)y=f\left(x\right)y=f(x)

y=ln⁡(x−4)y=\ln\left(x-4\right)y=ln(x−4)

2. Solve for xxx

ey=eln⁡(x−4)e^y=e^{\ln\left(x-4\right)}ey=eln(x−4)
ey=x−4e^y=x-4ey=x−4
ey+4=xe^y+4=xey+4=x
x=ey+4x=e^y+4x=ey+4

3. Swap xxx and yyy

y=ex+4y=e^x+4y=ex+4

4. Replace yyy with f−1(x)f^{-1}\left(x\right)f−1(x), this is the inverse function

Therefore, f−1(x)=ex+4f^{-1}\left(x\right)=e^x+4f−1(x)=ex+4

Domain and range:

𝑓 has domain (4,∞)\left(4,\infty\right)(4,∞) and range (−∞,∞)\left(-\infty,\infty\right)(−∞,∞)
f−1f^{-1}f−1 has domain (−∞,∞)\left(-\infty,\infty\right)(−∞,∞) and range (4,∞)\left(4,\infty\right)(4,∞)

Solve for xxx in the equation 2ln⁡x=ln⁡4+ln⁡(x−1)2\ln x=\ln4+\ln\left(x-1\right)2lnx=ln4+ln(x−1).