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Different Operations On Log Function Like Inverse Limits

Finding the Inverse of a Function & Limit of Expression involves Log Function, practice to solve inequalities containing logarithmic function and domain and range of log functions.

Domain and Range of Logarithmic Functions

  • Domain of \(log_ax\) is \(x>0\) or \((0, \infty)\). [ for all valued  \('a'\) ]
  • Range of \(log_ax\) is R or \((-\infty, \infty)\) (It takes all possible value)
  • \(log_ax=f(x)\) is a one to one function for all values of \('a'\).
  • To find the domain of \(f(x)=log_a\;(g(x))\), when \(g(x)\) is some expression in \(x\) , we solve the inequality \(g(x)>0\)

Illustration Questions

Find the domain of \(f(x)=log_5(x^2-16)\)

A \((-\infty, 2)\cup(5,\infty)\)

B \((-\infty, 6)\cup(10,\infty)\)

C \((-\infty, -4)\cup(4,\infty)\)

D \((2,7)\)

×

Domain of \(f(x)=log_a\Big(g(x)\Big)\) is \(g(x)>0\).

\(\therefore \;\)Domain of \(log_5\Big(x^2-16\Big)\) is \(x^2-16>0\)

 \(\Rightarrow (x-4)(x+4)>0\) 

Interval \(x+4\) \(x-4\) \((x+4)(x-4)\)
\(x>4\) + + +
\(-4<x<4\) +
\(x<-4\) +

\(\Rightarrow x\in(-\infty,-4)\cup(4,\infty)\)

\(\therefore \;\) Domain of \('f'\) is \((-\infty, -4)\cup(4,\infty)\).

Find the domain of \(f(x)=log_5(x^2-16)\)

A

\((-\infty, 2)\cup(5,\infty)\)

.

B

\((-\infty, 6)\cup(10,\infty)\)

C

\((-\infty, -4)\cup(4,\infty)\)

D

\((2,7)\)

Option C is Correct

Finding the Inverse of a Function whose Expression involves Log Function

  • To find the inverse of a function:
  1. Let \(y = f(x)\) be the given function.
  2. Solve \(x\) in term of \(y\).
  3. Interchange \(x\) and \(y\), the new \(y\) obtained is the required inverse.

Illustration Questions

Find the inverse function of the function  \(f(x)=log_5(6+log_5x)\).

A \(f^{-1}(x)=5^{(5^{x}-6)}\)

B \(f^{-1}(x)=6^{(5^{x}-6)}\)

C \(f^{-1}(x)=7^x\)

D \(f^{-1}(x)=\ell n\,x\)

×

\(f(x)=log_5(6+log_5x)\)

Step 1 : 

\(y=log_5(6+log_5x)\)

Step 2 : 

\(5^y=log_5\,x+6\Rightarrow log_5\,x=5^y-6\)

\(x=5^{(5^y-6)}\)

 

 

Step 3 : 

\(y=5^{(5^x-6)}=f^{-1}(x)\)

 

 

Find the inverse function of the function  \(f(x)=log_5(6+log_5x)\).

A

\(f^{-1}(x)=5^{(5^{x}-6)}\)

.

B

\(f^{-1}(x)=6^{(5^{x}-6)}\)

C

\(f^{-1}(x)=7^x\)

D

\(f^{-1}(x)=\ell n\,x\)

Option A is Correct

Finding Limit of Expression that involves Log Function

If \(a>1\), graph of \(\,log_ax\) is

 

  • Observe that,

 \(\lim\limits_{x\rightarrow\infty}\,log_ax=\infty\) ...(1)

\(\lim\limits_{x\rightarrow0^+}\,log_ax=-\infty\) ...(2)

  • This means that log of very large quantities take very large values and log of small positive quantities take very large negative values.

If \(0<a<1\), graph of \(log_ax\) is:

  • Observe that

 \(\lim\limits_{x\rightarrow\infty}\,log_ax=-\infty\) ...(3)

\(\lim\limits_{x\rightarrow0^+}\,log_ax=\infty\) ...(4)

  • This means that log of very large positive quantities take very large negative values and log of small positive quantities take very large positive values.

Illustration Questions

Evaluate   \(\lim\limits_{x\rightarrow3^+}\,log_2\,(5x-15)\).  

A \(\dfrac {5}{3}\)

B 0

C \(-\infty\)

D \(\infty\)

×

Here, \(a=2,\;5x-15\) will take small positive values where \(x\)take \(3^+\)i.e. values just larger than 3.

\(\therefore\;\lim\limits_{x\rightarrow3^+}\,log_2\,(5x-15)\)

\(=\lim\limits_{x\rightarrow3^+}\,log_2\,y\) (where \(y\) is small positive )

\(=-\infty\) (by (2))

Evaluate   \(\lim\limits_{x\rightarrow3^+}\,log_2\,(5x-15)\).  

A

\(\dfrac {5}{3}\)

.

B

0

C

\(-\infty\)

D

\(\infty\)

Option C is Correct

More Limits on Log Function (infinity minus infinity\(\infty -\infty\) form)

  • To evaluate limits of the form \(\lim\limits_{x\rightarrow \infty}\Big(log\,f(x)-log\,g(x)\Big)\), where \(f(x)\) and \(g(x)\) are both tending to \(\infty\) (\(\infty -\infty\) form) we first use properties of log to get \(\lim\limits_{x\rightarrow \infty}\,log\left ( \dfrac {f(x)}{g(x)}\right)\).
  • Then according to the degree of  \(f(x)\) and \(g(x)\) ,we give the answer.

 

Illustration Questions

Evaluate  \(\lim\limits_{x\rightarrow \infty}\Big(\ell n\,(x+7)-\ell n\,(x+5)\Big)\).

A \(\dfrac {1}{2}\)

B \(-7\)

C 5

D \(0\)

×

\(\lim\limits_{x\rightarrow \infty}\Big(\ell n\,(x+7)-\ell n\,(x+5)\Big)\)

\(=\lim\limits_{x\rightarrow \infty}\dfrac {(x+7)}{(x+5)}\)      (Use property of log)

\(=\lim\limits_{x\rightarrow \infty}\dfrac {\Big(1+\dfrac {7}{x}\Big)} {\Big(1+\dfrac {5}{x}\Big)}=\ell \,n\,1=0\)    \(\left (\because\dfrac {7}{x}\,,\dfrac {5}{x} \rightarrow 0 \right)\)

Evaluate  \(\lim\limits_{x\rightarrow \infty}\Big(\ell n\,(x+7)-\ell n\,(x+5)\Big)\).

A

\(\dfrac {1}{2}\)

.

B

\(-7\)

C

5

D

\(0\)

Option D is Correct

Solving Inequalities Containing Log Function

  • If \(a > 1\), then
  1. \(log_ax>b\Rightarrow x>a^b \Rightarrow x\in(a^b, \infty)\)
  2. \(log_ax<b\Rightarrow x<a^b \Rightarrow x\in(0, a^b)\) because \(x\) has to be positive for its log to be defined.

e.g.

  1. \(log_2x>3\Rightarrow x>2^3 \text { or } x > 8\)
  2. \(log_{1/2}x>3\Rightarrow x<\Big(\dfrac {1}{2}\Big)^3 \text { or } x < \dfrac {1}{8}\) .

 In (1) base 2 is greater than 1  whereas in (2) base \(\dfrac {1}{2}\)  is less than 1 .

Illustration Questions

Solve for \(x\) the inequality \(2-5\,\ell n\,x>7\).

A \(x\in\left (\dfrac {1}{e},\infty\right)\)

B \(x\in(1,e)\)

C \(x\in\left (0,\;\dfrac {1}{e}\right)\)

D \(x\in(e, \infty)\)

×

\(2-5\,\ell n\,x>7\;\)

\(\Rightarrow 5\,\ell n\,x<2-7\)

\(\Rightarrow \ell n\,x<\dfrac {-5}{5}\;\)

\(\Rightarrow \ell n\,x<-1\)

\(\therefore \;x<e^{-1} \;\) 

\(\Rightarrow x< \dfrac {1}{e}\)

\(\Rightarrow x\in\left ( 0,\dfrac {1}{e} \right)\)      because \(x\)has to be positive.

Solve for \(x\) the inequality \(2-5\,\ell n\,x>7\).

A

\(x\in\left (\dfrac {1}{e},\infty\right)\)

.

B

\(x\in(1,e)\)

C

\(x\in\left (0,\;\dfrac {1}{e}\right)\)

D

\(x\in(e, \infty)\)

Option C is Correct

Graph of logax v/s x

  • If  \(a>1\) then \(log_ax\) is an increasing function of \(x\).

  • If  \(0<a<1\) then \(log_ax\) is a decreasing function of \(x\).

 

  • The point \((1, 0)\) is on every \(y=log_ax\) curve, for all allowed values of \('a'\).
  • For  \(a>1\), the graph of  \(log_ax\) for different values of \('a'\) will be as shown in figure.

  • As 'a' increases the rate of increase of \(log\) function decreases.

 

  • If  \(0<a<1\) , then the graph of \(log_ax\) for different values of \('a'\) will be as shown in figure.

 

Illustration Questions

Which of the following graphs shows the correct sequence of \(log_ax\) graphs?

A

B

C

D

×

If a > 1,  \(log_ax\) increases, the rate of increase decreases with increasing values of \('a'\).

\(\therefore\) (A) is correct option.

 

Which of the following graphs shows the correct sequence of \(log_ax\) graphs?

A image
B image
C image
D image

Option A is Correct

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