Informative line

Log Function And Its Equation

Learn how to solve elementary equations containing exponential functions based on definition. Practice logarithmic functions calculus and properties of log x graph functions.

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 an 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

Logarithmic Functions

Let \(f(x)=a^x\), we define  \(f^{-1}(x)=log_a\,x\)  as the inverse function of \(f\). It is called the logarithmic function with base \('a'\).

\(log_a\,x=y\;\iff\;a^y=x\)

  • \(log_a\,x\)  is the  power of  \('a'\)  required to reach  \(x\).
  • \(f\) \(log_2\,8=3\)  because \(2^3=8\)  (It is the power of 2 required to reach 8, it is 3)

Illustration Questions

Evaluate \(log_3\,27\).

A 4

B 3

C 2

D –1

×

\(log_3\,27=3\)  because \(3^3=27\)

Evaluate \(log_3\,27\).

A

4

.

B

3

C

2

D

–1

Option B is Correct

Properties of Logarithmic Function

  1. If  \(x,\,y>0\)  then  \(log_{a}\,(xy)=log_{a}\,x+log_{a}\,y\)  &  \(a>1\)
  2. If  \(x,\,y>0\)  then  \(log_{a}\left(\dfrac{x}{y}\right)=log_{a}\,x-log_{a}\,y\)  &  \(a>1\)
  3. If  \(x,\,y>0\)  then  \(log_{a}\,(x^r)=r\,log_{a}\,x\)

We can write, \(log_{10}\,6=log_{10}\,2×3=log_{10}\,2+log_{10}\,3\)

and  \(log_{10}\dfrac{10}{3}=log_{10}\,10-log_{10}\,3=1-log_{10}\,3\)

and  \(log_{10}\,25=log_{10}\,5^2=2log_{10}\,5\)

  • Logarithm functions convert products into sum and ratios into difference.

Illustration Questions

Find the value of  \(log_{2}\,12+log_{2}\,6-log_{2}\,9\).

A \(log_{2}\,3\)

B –1

C 5

D 3

×

Use, \(log_{a}\,x+log_{a}\,y=log_{a}\,xy\)

\(log_{2}\,12+log_{2}\,6=log_{2}\,72\)

Use,  \(log_{a}\,x-log_{a}\,y=log_{a}\dfrac{x}{y}\)

\(\therefore\,(log_{6}\,12+log_{2}\,6)-(log_{2}\,9)\)

\(=log_{2}\,72-log_{2}\,9\)

\(=log_{2}\dfrac{72}{9}=log_{2}\,8\)

\(=3\)

Find the value of  \(log_{2}\,12+log_{2}\,6-log_{2}\,9\).

A

\(log_{2}\,3\)

.

B

–1

C

5

D

3

Option D is Correct

Expressing many Log Quantities as one Log Quantity

Properties of Log

  1. If \(x,\,y>0\) & a >1 then \(log_{a}\,(xy)=log_{a}\,x+log_{a}\,y\)
  2. If \(x,\,y>0\) & a >1 then \(log_{a}\left(\dfrac{x}{y}\right)=log_{a}\,x-log_{a}\,y\)
  3. If \(x,\,y>0\) & a >1 then \(log_{a}\,(x^r)=r\,log_{a}\,x\)

 

Sometimes the reverse of above properties will be used in the problem.

e.g.

\(5\,log_{10}\,x^2-2\,log\,y+7\,log\dfrac{5}{y}\)

can be written as,

\(log_{10}\,x^{10}-log\,y^2+log\,\left(\dfrac{5}{y}\right)^7\)

\(=log_{10}\,\dfrac{x^{10}}{y^2}×\left(\dfrac{5}{y}\right)^7\)

\(=log_{10}\,\dfrac{x^{10}\,5^7}{y^9}\)

Illustration Questions

Express the following quantity as simple logarithm: \(2\,log_{10}\,x-3\,log_{10}\,y+log\left(\dfrac{x}{y}\right)^2\)

A \(log_{10}\,(x^2y^3)\)

B \(log_{10}\,\left(\dfrac{y^2}{x^3}\right)\)

C \(log_{10}\left(\dfrac{x^4}{y^5}\right)\)

D \(log_{10}\left(\dfrac{y^5}{x^4}\right)\)

×

\(2\,log_{10}\,x=log_{10}\,x^2\)

\(3\,log_{10}\,y=log_{10}\,y^3\)

\(\therefore\) Given quantity \(=log_{10}\,x^2-log_{10}\,y^3+log\left(\dfrac{x}{y}\right)^2\)

\(=log_{10}\dfrac{x^2}{y^3}+log_{10}\left(\dfrac{x}{y}\right)^2\)

\(=log_{10}\left(\dfrac{x^2}{y^3}×\dfrac{x^2}{y^2}\right)\)

\(=log_{10}\left(\dfrac{x^4}{y^5}\right)\)

Express the following quantity as simple logarithm: \(2\,log_{10}\,x-3\,log_{10}\,y+log\left(\dfrac{x}{y}\right)^2\)

A

\(log_{10}\,(x^2y^3)\)

.

B

\(log_{10}\,\left(\dfrac{y^2}{x^3}\right)\)

C

\(log_{10}\left(\dfrac{x^4}{y^5}\right)\)

D

\(log_{10}\left(\dfrac{y^5}{x^4}\right)\)

Option C is Correct

Expanding Single Logarithm Quantity

  1. If \(x,\,y>0\) & a > 1 then \(log_{a}\,(xy)=log_{a}\,x+log_{a}\,y\)
  2. If \(x,\,y>0\) & a > 1 then \(log_{a}\left(\dfrac{x}{y}\right)=log_{a}\,x-log_{a}\,y\)
  3. If \(x,\,y>0\) & a > 1 then \(log_{a}\,(x^r)=r\,log_{a}\,x\)

 

A single logarithmic function which contain complicated expression can be simplified to many \(log\) term by using the above properties.

Example:   \(log_{10}\,3\sqrt{\dfrac{y+3}{2x+7}}\) can be written as,

\(=log_{10}\left(\dfrac{y+3}{2x+7}\right)^{1/3}\)

\(=\dfrac{1}{3}log_{10}\left(\dfrac{y+3}{2x+7}\right)\)

\(=\dfrac{1}{3}\left[log_{10}(y+3)-log_{10}(2x+7)\right]\)

Illustration Questions

Use properties of logarithm to expand \(log_{10}\,\left(\sqrt{\dfrac{x^2+1}{x^2+3}}\right)\)

A \(\dfrac{1}{2}[log_{10}\,(x^2+1)-log_{10}\,(x^2+3)]\)

B \(\dfrac{1}{2}log\,\left(\dfrac{x^2+3}{x^2+1}\right)\)

C \(\dfrac{1}{2}[log_{10}\,(x^2+4)-log_{10}\,x]\)

D \(2\,log\,x^2\)

×

\(log_{10}\,\left(\sqrt{\dfrac{x^2+1}{x^2+3}}\right)=\dfrac{1}{2}log_{10}\,\left(\dfrac{x^2+1}{x^2+3}\right)\)

\(=\dfrac{1}{2}\left[log_{10}\,(x^2+1)-log_{10}\,(x^2+3)\right]\)

Use properties of logarithm to expand \(log_{10}\,\left(\sqrt{\dfrac{x^2+1}{x^2+3}}\right)\)

A

\(\dfrac{1}{2}[log_{10}\,(x^2+1)-log_{10}\,(x^2+3)]\)

.

B

\(\dfrac{1}{2}log\,\left(\dfrac{x^2+3}{x^2+1}\right)\)

C

\(\dfrac{1}{2}[log_{10}\,(x^2+4)-log_{10}\,x]\)

D

\(2\,log\,x^2\)

Option A is Correct

Solving Elementary Logarithmic Equations (Based on Definition)

If we need to solve a simple logarithmic equation often, the definition of \(log\) will be required to convert that equation into simple linear one.

Example:  Solve \(log_{2}\,(3x-5)=3\)

\(\Rightarrow\,3x-5=2^3\)           \((\log_{a}\,x=y\Rightarrow\,x=a^y)\)

\(\Rightarrow\,3x=5+8\)

\(\Rightarrow\,x=\dfrac{13}{3}\)

Illustration Questions

Solve for \(x\), the equation \(log_{10}\,(5x-2)=2\) .

A \(x=\dfrac{102}{5}\)

B \(x=\dfrac{5}{7}\)

C \(x=-\dfrac{2}{3}\)

D \(x=\dfrac{2}{105}\)

×

\(log_{10}\,(5x-2)=2\)

\(\Rightarrow\,5x-2=10^2\)

\(\Rightarrow\,5x-2=100\)

\(\Rightarrow\,5x=102\)

\(\Rightarrow\,x=\dfrac{102}{5}\)

Solve for \(x\), the equation \(log_{10}\,(5x-2)=2\) .

A

\(x=\dfrac{102}{5}\)

.

B

\(x=\dfrac{5}{7}\)

C

\(x=-\dfrac{2}{3}\)

D

\(x=\dfrac{2}{105}\)

Option A is Correct

Solving Elementary Equations containing Exponential Functions

If the given equation contains exponential function in which the exponents contain variable \(x\) then to solve for \(x\), we take \(log\) on both sides to the same base whose power is raised. This will convert the equation to simple equation.

e.g.  Solve  \(3^{2x-7}=5\)

Take \(log\) , both sides to base 3

\(\Rightarrow\,(2x-7)\log_{3}\,3=\log_{3}\,5\)

\(\Rightarrow\,2x-7=\log_{3}\,5\)

\(\Rightarrow\,x=\dfrac{7+\log_{3}\,5}{2}\)

Illustration Questions

Solve for \(x\), the equation \(5^{2x-3}=4\) .

A \(x=\dfrac{(\log_{5}\,4+3)}{2}\)

B \(x=\log_{4}\,5\)

C \(x=\dfrac{(log_{4}\,5+11)}{9}\)

D \(x=e^2\)

×

\(5^{2x-3}=4\to\) Take \(log\), both sides to the base 5

\(\Rightarrow\,log_{5}\,\left(5^{2x-3}\right)=\log_{5}\,4\)

\(\Rightarrow\,(2x-3)\,log_{5}\,5=log_{5}\,4\)

\(\Rightarrow\,2x-3=log_{5}\,4\)

\(\Rightarrow\,x=\dfrac{log_{5}\,4+3}{2}\)

Solve for \(x\), the equation \(5^{2x-3}=4\) .

A

\(x=\dfrac{(\log_{5}\,4+3)}{2}\)

.

B

\(x=\log_{4}\,5\)

C

\(x=\dfrac{(log_{4}\,5+11)}{9}\)

D

\(x=e^2\)

Option A is Correct

The Natural Logarithm (log to the base e)

  • \(log_{e}\,x=ln\,x\to\) (called natural logarithm)

 \(ln\) stands for  \(log\)  natural.

  • \(ln\,x=y\)
  • \(\Rightarrow\,x=e^y\)

Illustration Questions

Solve the equation \(e^{2x}+e^x-6=0\) for \(x\) .

A \(x=ln\,2\)

B \(x=ln\,3\)

C \(x=ln\,6\)

D \(x=ln\,500\)

×

\(e^{2x}+e^x-6=0\)

put \(e^x=t\) ,

\(\Rightarrow\,t^2+t-6=0\)

\(\Rightarrow\,(t+3)\,(t-2)=0\)

\(\Rightarrow\,t=-3,\;t=2\)

\(\Rightarrow\,e^x=-3 ,\;e^x=2\)

\(\Rightarrow\,e^x=-3\)    (Not possible because \(e^x>0\,\forall\,x\))

\(\therefore\,e^x=2\)

 Take \(log\) , both sides to base \(e\)

\(\Rightarrow\,x=ln\,2\) 

Solve the equation \(e^{2x}+e^x-6=0\) for \(x\) .

A

\(x=ln\,2\)

.

B

\(x=ln\,3\)

C

\(x=ln\,6\)

D

\(x=ln\,500\)

Option A is Correct

Practice Now