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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.  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 B C D 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)

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

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,

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

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

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

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 $$log_{2}\,(3x-5)=3$$

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

$$\Rightarrow\,3x=5+8$$

$$\Rightarrow\,x=\dfrac{13}{3}$$

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

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.

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,

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

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

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

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

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

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

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

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

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.

• $$ln\,x=y$$
• $$\Rightarrow\,x=e^y$$

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