Some important Logarithmic and Exponential formulae

1. If $\mathrm{a}^{\mathrm{x}}=\mathrm{y}$, then $\mathrm{x}=\log _{\mathrm{a}} \mathrm{y}$

2. $\log _{\mathrm{a}} \mathrm{a}=1$ & $\log _{\mathrm{a}} 1=0$

3. $\mathrm{a}^{\log _{a} \mathrm{n}}=\mathrm{n}$

4. $\log _{\mathrm{a}} \mathrm{mn}=\log _{\mathrm{a}} \mathrm{m}+\log _{\mathrm{a}} \mathrm{n}$

5. $\log _{a}\left(\dfrac{m}{n}\right)=\log _{a} m-\log _{a} n$

6. $\log _{\mathrm{a}} \mathrm{m}^{\mathrm{n}}=\mathrm{n} \log _{\mathrm{a}} \mathrm{m}$

7. $\log _{\mathrm{b}} \mathrm{a}=\dfrac{\log _{\mathrm{c}} \mathrm{a}}{\log _{\mathrm{c}} \mathrm{b}}$

8. $\log _{a^{n}} a^{m}=\dfrac{m}{n}$

9. $\log _{\mathrm{b}} \mathrm{a}=\dfrac{1}{\log _{\mathrm{a}} \mathrm{b}}$

10. $\log _{e}(1+x)=x-\dfrac{x^{2}}{2}+\dfrac{x^{3}}{3}-\dfrac{x^{4}}{4}+\ldots \ldots \ldots .$

11. $\log _{e}(1-x)=-x-\dfrac{x^{2}}{2}-\dfrac{x^{3}}{3}-\dfrac{x^{4}}{4}\ldots \ldots \ldots .$

12. $\log _{e}(1+x)(1-x)=\log _{e}\left(1-x^{2}\right)=-2\left(\dfrac{x^{2}}{2}+\dfrac{x^{4}}{4}+\dfrac{x^{6}}{6}+\ldots \ldots \ldots \ldots . ..\right)$

13. $\log _{e}\left(\dfrac{1+x}{1-x}\right)=2\left(x+\dfrac{x^{3}}{3}+\dfrac{x^{5}}{5}+\ldots \ldots \ldots \ldots \ldots \ldots . . . .\right)$

14. $\log _{\mathrm{e}} 2=1-\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\ldots \ldots \ldots \ldots .$

15. $\mathrm{e}^{\mathrm{x}}=1+\dfrac{\mathrm{x}}{1 !}+\dfrac{\mathrm{x}^{2}}{2 !}+\dfrac{\mathrm{x}^{3}}{3 !}+\ldots \ldots \ldots \ldots .$

16. $e^{x}+e^{-x}=2\left(1+\dfrac{x^{2}}{2 !}+\dfrac{x^{4}}{4 !}+\ldots \ldots \ldots \ldots .\right)$

17. $e^{x}-e^{-x}=2\left(\dfrac{x}{1 !}+\dfrac{x^{3}}{3 !}+\dfrac{x^{5}}{5 !} \ldots \ldots \ldots \ldots ..\right)$

18. $\mathrm{e}=1+\dfrac{1}{1 !}+\dfrac{1}{2 !}+\dfrac{1}{3 !}+\ldots \ldots \ldots \ldots .$

$\quad \mathrm{e}$ is an irrational number & it lies between $2 & 3 . \mathrm{e} \cong 2.7183$

19. $a^y=e^{y \log _c a}=1+y\left(\log _c a\right)+\dfrac{y^2}{2 !}\left(\log _e a\right)^2+\dfrac{y^3}{3 !}\left(\log _c a\right)^3+\ldots \ldots \ldots \ldots \ldots.$

20. $\log _{2} \log _{2} \sqrt{\sqrt{\sqrt{\sqrt{\ldots \ldots \ldots \ldots \ldots . . \sqrt{2}}}}}=-\mathrm{n}$ when $\mathrm{n}$ is the number of square roots

21. $a^{\sqrt{\log _{a} b}}=b^{\sqrt{\log _{b} a}}$

Vn-Method

(i) To find the sum of series of the form

$\dfrac{1}{a _{1} a _{2} \ldots \ldots \ldots . a _{r}}+\dfrac{1}{a _{2} a _{3} \ldots \ldots \ldots \ldots a _{r+1}}+\ldots \ldots \ldots \ldots \ldots \ldots+\dfrac{1}{a _{n} a _{n+1} \ldots \ldots \ldots \ldots a _{n+r-1}}$ where $a _{1}, a _{2}$,

$\left(\text { Here } T_n=\dfrac{1}{a_n a_{n+1} \ldots \ldots \ldots \ldots a_{n+r-1}}\right)$

………are in A.P. Let $\mathrm{V} _{\mathrm{n}}=\dfrac{1}{\mathrm{a} _{\mathrm{n}+1} \mathrm{a} _{\mathrm{n}+2} \ldots \ldots \ldots \ldots \mathrm{a} _{\mathrm{n}+\mathrm{r}-1}}\left(\right.$ avoiding first term for $\mathrm{V} _{\mathrm{n}}$ ie $\mathrm{a} _{\mathrm{n}}$ in $T _{\mathrm{n}}$ )

$\mathrm{T} _{\mathrm{n}}=\dfrac{-1}{\mathrm{~d}(\mathrm{r}-1)}\left(\mathrm{V} _{\mathrm{n}}-\mathrm{V} _{\mathrm{n}-1}\right)=\dfrac{1}{\mathrm{~d}(\mathrm{r}-1)}\left(\mathrm{V} _{\mathrm{n}-1}-\mathrm{V} _{\mathrm{n}}\right)$

put $\mathrm{n}=1,2,3 \ldots \ldots \ldots \ldots . . \mathrm{n}$ and add to get $\mathrm{S} _{\mathrm{n}}$.

$\mathrm{S} _{\mathrm{n}}=\mathrm{T} _{1}+\mathrm{T} _{2}+\ldots \ldots . \mathrm{T} _{\mathrm{n}}=\dfrac{1}{\mathrm{~d}(\mathrm{r}-1)}\left(\mathrm{V} _{0}-\mathrm{V} _{\mathrm{n}}\right)$

$$ =\dfrac{1}{d(r-1)}\left(\dfrac{1}{a _{1} a _{2} \cdots \ldots . a _{r-1}}-\dfrac{1}{a _{n+1} a _{n+2} \cdots \ldots . a _{n+r-1}}\right) $$

Eg:

(a) $\dfrac{1}{1.2 .3}+\dfrac{1}{2.3 .4}+\ldots \ldots \ldots \ldots . .+\dfrac{1}{\mathrm{n}(\mathrm{n}+1)(\mathrm{n}+2)}=\dfrac{1}{1 .(3-1)}\left(\dfrac{1}{1.2}-\dfrac{1}{(\mathrm{n}+1)(\mathrm{n}+2)}\right)$

$\quad=\dfrac{1}{4}-\dfrac{1}{2(\mathrm{n}+1)(\mathrm{n}+2)}$

(b) $\dfrac{1}{1.2 .3 .4}+\dfrac{1}{2 \cdot 3 \cdot 4.5}+\ldots \ldots \ldots . . \dfrac{1}{n(n+1)(n+2)(n+3)}$

$\quad=\dfrac{1}{1 .(4-1)}\left(\dfrac{1}{1.2 \cdot 3}-\dfrac{1}{(\mathrm{n}+1)(\mathrm{n}+2)(\mathrm{n}+3)}\right)$

(ii) Summation of series of the form $\mathrm{a} _{1} \mathrm{a} _{2} \ldots \ldots \mathrm{a} _{\mathrm{r}}+\mathrm{a} _{2} \mathrm{a} _{3} \ldots \ldots \ldots \mathrm{a} _{\mathrm{r}+1}+$. $\mathrm{a} _{1}, \mathrm{a} _{2} \ldots \ldots$. are in A.P

Here $T _{n}=a _{n} a _{n+1} \cdots \cdots \ldots a _{n+r-1}$

Let $V _{n}{ } _{n}=a _{n} a _{n+1} \cdots \cdots \cdots \ldots \ldots . . . a _{n-r+1} a _{n-r}\left(\right.$ Take one term extra in $T _{n}$ for $\left.V _{n}\right)$

$\mathrm{T} _{\mathrm{n}}=\dfrac{1}{(\mathrm{r}+1) \mathrm{d}}\left(\mathrm{V} _{\mathrm{n}}-\mathrm{V} _{\mathrm{n}-1}\right)$.

Put $\mathrm{n}=1,2,3 \ldots \ldots .$. and add to get $\mathrm{S} _{\mathrm{n}}$

$\mathrm{S} _{\mathrm{n}}=\mathrm{T} _{1}+\mathrm{T} _{2}+\ldots \ldots \ldots \ldots+\mathrm{T} _{\mathrm{n}}=\dfrac{1}{\mathrm{~d}(\mathrm{r}+1)}\left(\mathrm{V} _{\mathrm{n}}-\mathrm{V} _{0}\right)=\dfrac{1}{\mathrm{~d}(\mathrm{r}+1)}\left(\mathrm{a} _{\mathrm{n}} \mathrm{a} _{\mathrm{n}+1} \ldots \ldots \ldots . \mathrm{a} _{\mathrm{n}+\mathrm{r}}-\mathrm{a} _{0} \mathrm{a} _{1} \mathrm{a} _{2} \ldots \ldots \ldots \ldots \mathrm{a} _{\mathrm{r}}\right)$

where $\mathrm{a} _{0}=\mathrm{a} _{1}-\mathrm{d}$

Eg

(a) $1.2+2.3+\ldots \ldots \ldots \ldots \ldots \ldots \ldots+n(\mathrm{n}+1)=\dfrac{1}{1 .(2+1)}((\mathrm{n}+1)(\mathrm{n}+2)-0.1 .2)=\dfrac{\mathrm{n}(\mathrm{n}+1)(\mathrm{n}+2)}{3}$

(b) $1.2 .3 .4+2.3 .4 .5+$ $+\mathrm{n}(\mathrm{n}+1)(\mathrm{n}+2)(\mathrm{n}+3)$

$\quad\begin{aligned} & =\dfrac{1}{1 .(4+1)}(\mathrm{n}(\mathrm{n}+1)(\mathrm{n}+2)(\mathrm{n}+3)(\mathrm{n}+4)-0.1 \cdot 2 \cdot 3 \cdot 4) \\ \\ & =\dfrac{1}{5} \mathrm{n}(\mathrm{n}+1)(\mathrm{n}+2)(\mathrm{n}+3)(\mathrm{n}+4) \end{aligned}$

SOLVED EXAMPLES

1. If the sides of a triangle are in A.P and the greatest angle of the triangle is double the smallest, then the ratio of sides of the triangle is

(a) $3: 4: 5$

(b) $4: 5: 6$

(c) $5: 6: 7$

(d) None of these

Show Answer

Solution :

Applying sine rule, we have

$\dfrac{\mathrm{a}-\mathrm{d}}{\sin \theta}=\dfrac{\mathrm{a}}{\sin (\pi-3 \theta)}=\dfrac{\mathrm{a}+\mathrm{d}}{\sin 2 \theta}$

$\Rightarrow \dfrac{\mathrm{a}-\mathrm{d}}{\sin \theta}=\dfrac{\mathrm{a}}{3 \sin \theta-4 \sin ^{3} \theta}=\dfrac{\mathrm{a}+\mathrm{d}}{2 \sin \theta \cos \theta}$

$\Rightarrow \dfrac{\mathrm{a}-\mathrm{d}}{1}=\dfrac{\mathrm{a}}{3-4 \sin ^{2} \theta}=\dfrac{\mathrm{a}+\mathrm{d}}{2 \cos \theta}$ gives $\cos \theta=\dfrac{\mathrm{a}+\mathrm{d}}{2(\mathrm{a}-\mathrm{d})}$

Also, $3-4 \sin ^{2} \theta=\dfrac{\mathrm{a}}{\mathrm{a}-\mathrm{d}}$

$\Rightarrow 3-4\left(1-\cos ^{2} \theta\right)=\dfrac{a}{a-d}$ gives $-1+\left(\dfrac{a+d}{a-d}\right)^{2}=\dfrac{a}{a-d}$

$\Rightarrow \dfrac{4 \mathrm{ad}}{\mathrm{a}-\mathrm{d}}=\mathrm{a} \Rightarrow 4 \mathrm{ad}=\mathrm{a}^{2}-\mathrm{ad}$ gives $\mathrm{a}=5 \mathrm{~d}$

$\therefore$ sides $\mathrm{a}-\mathrm{d}: \mathrm{a}: \mathrm{a}+\mathrm{d}$

$\quad \quad \quad 5\mathrm{~d}-\mathrm{d}: 5 \mathrm{~d}: 5 \mathrm{~d}+\mathrm{d}=4: 5: 6$

Answer: (b)

2. The sum of

$\dfrac{3}{1.2} \cdot \dfrac{1}{2}+\dfrac{4}{2.3}\left(\dfrac{1}{2}\right)^{2}+\dfrac{5}{3.4} \cdot\left(\dfrac{1}{2}\right)^{3}+$……………$\mathrm{n}$ terms is

(a) $1-\dfrac{1}{(\mathrm{n}+1) 2^{\mathrm{n}}}$

(b) $1-\dfrac{1}{\mathrm{n} \cdot 2^{\mathrm{n}-1}}$

(c) $1+\dfrac{1}{(\mathrm{n}+1) 2^{\mathrm{n}}}$

(d) None of these

3. Coefficient of $x^{49}$ in the expansion of $(x-1)(x-3)(x-5)…………….(x-99)$ is

(a) $-99^{2}$

(b) 1

(c) -2500

(d) None of these

4. The coefficients of $x^{15}$ in the product

$(1-\mathrm{x})(1-2 \mathrm{x})\left(1-2^{2} \mathrm{x}\right) \ldots \ldots \ldots \ldots . .\left(1-2^{15} \mathrm{x}\right)$ is

(a) $2^{105}-2^{121}$

(b) $2^{121}-2^{105}$

(c) $2^{120}-2^{104}$

(d) None of these

5. The sum to $2 \mathrm{n}$ terms of the series

$\dfrac{3}{4}+\dfrac{7}{4}+\dfrac{15}{16}+\dfrac{31}{16}+\dfrac{63}{64}+\dfrac{127}{64}+$………………is

(a) $3 \mathrm{n}-\dfrac{2}{3}\left(1-\dfrac{1}{4^{\mathrm{n}}}\right)$

(b) $3 \mathrm{n}-\dfrac{10}{21}\left(1-\dfrac{1}{4^{n}}\right)$

(c) $3 n-\dfrac{13}{21} \dfrac{1}{4^{n}}$

(d) None of these

6. The sum to $n$ terms of the series

$\dfrac{1}{1-\dfrac{1}{4}}+\dfrac{1}{(1+3)-\dfrac{1}{4}}+\dfrac{1}{(1+3+5)-\dfrac{1}{4}}+\ldots \ldots \ldots \infty$

(a) $\dfrac{2 \mathrm{n}}{2 \mathrm{n}+1}$

(b) $\dfrac{4 \mathrm{n}}{2 \mathrm{n}+1}$

(c) $\dfrac{2}{2 n+1}$

(d) None

7. $\quad$ The sum of the series

$\dfrac{1}{1.3}+\dfrac{2}{1.3 .5}+\dfrac{3}{1.3 .5 .7}+$……………….$\infty$ is

(a) 1

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

(c) $\dfrac{3}{2}$

(d) None

Exercise

1.* Let $S _{1}, S _{2}, \ldots \ldots \ldots \ldots \ldots . .$. . be squares such that for each $n \geq 1$ the length of a side of $S _{n}$ equals the length of a diagonal of $\mathrm{S} _{\mathrm{n}+1}$. If the length of a side of $\mathrm{S} _{1}$ is $10 \mathrm{~cm}$, then for which of the following values of $\mathrm{n}$ is the area of $\mathrm{S} _{\mathrm{n}}$ less than $1 \mathrm{sq} . \mathrm{cm}$.

(a) 7

(b) 8

(c) 9

(d) 10

2. If $\mathrm{a}, \mathrm{b}, \mathrm{c}$ are is $\mathrm{A} . \mathrm{P}$ and $\mathrm{a}^{2}, \mathrm{~b}^{2}, \mathrm{c}^{2}$ are in H.P, then $\mathrm{b}^{2}=$

(a) $\dfrac{\mathrm{ca}}{2}$

(b) $2 \mathrm{ca}$

(c) $\dfrac{-\mathrm{ca}}{2}$

(d) $-2 \mathrm{ca}$.

3. Let the $\mathrm{HM}$ & $\mathrm{GM}$ of two positive numbers $\mathrm{a}$ & $\mathrm{~b}$ be in the ratio $4: 5$ then $\mathrm{a}: \mathrm{b}$ is

(a) $1: 2$

(b) $2: 3$

(c) $3: 4$

(d) $1: 4$

4. If $\cos (x-y), \cos x, \cos (x+y)$ are in H.P., then the value of $\cos x \sec \ddfrac{y}{2}$ is

(a) $\pm 1$

(b) $\pm \dfrac{1}{\sqrt{2}}$

(c) $\pm \sqrt{2}$

(d) $\pm \sqrt{3}$

5. If $x$ & $y$ are positive real numbers and $m, n$ are positive integers, then the minimum value of $\dfrac{x^{m} y^{n}}{\left(1+x^{2 m}\right)\left(1+y^{2 n}\right)}$ is

(a) 2

(b) $\dfrac{1}{4}$

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

(d) 1

6. There are two numbers $a \& b$ whose product is 192 and the quotient of A.M. by H.M. of their greatest common divisor and least common multiple is $\dfrac{169}{48}$. The smaller of $a$ & $b$ is

(a) 2

(b) 4

(c) 6

(d) 12

7. Consider the sequence 1, 2, 2, 3, 3, 3, where n occurs $n$ times. The number that occurs as $2007^{\text {th }}$ term is

(a) 61

(b) 62

(c) 63

(d) 64

8. Read the following paragraph and answer the questions.

Let $A _{1}, G _{1}, H _{1}$ denote the A.M., G.M., H.M of two distinct positive numbers. For $n \geq 2$, let $A _{n-1}$ and $H _{n-1}$ have A.M., G.M., H.M as $A _{n}, G _{n}, H _{n}$ respectively.

(i) Which of the following statements is correct?

(a) $\mathrm{G} _{1}>\mathrm{G} _{2}>\mathrm{G} _{3}>$…………….

(b) $\mathrm{G} _{1}<\mathrm{G} _{2}<\mathrm{G} _{3}<$…………….

(c) $\mathrm{G} _{1}=\mathrm{G} _{2}=\mathrm{G} _{3}=$…………….

(d) $\mathrm{G} _{1}<\mathrm{G} _{3}<\mathrm{G} _{5}$………………and $\mathrm{G} _{2}>\mathrm{G} _{4}>\mathrm{G} _{6}>\ldots$.

(ii) Which are of the following statement is correct?.

(a) $\mathrm{A} _{1}>\mathrm{A} _{2}>\mathrm{A} _{3}>$……………….

(b) $\mathrm{A} _{1}<\mathrm{A} _{2}<\mathrm{A} _{3}<$………………

(c) $\mathrm{A} _{1}>\mathrm{A} _{3}>\mathrm{A} _{5}>$………………and $\mathrm{A} _{2}<\mathrm{A} _{4}<\mathrm{A} _{6}<$………………

(d) $\mathrm{A} _{1}<\mathrm{A} _{3}<\mathrm{A} _{5}<$………………and $\mathrm{A} _{2}>\mathrm{A} _{4}>\mathrm{A} _{6}>$………………

(iii) Which are of the following statement is correct?

(a) $\mathrm{H} _{1}>\mathrm{H} _{2}>\mathrm{H} _{3}>$………………

(b) $\mathrm{H} _{1}<\mathrm{H} _{2}<\mathrm{H} _{3}<$………………

(c) $\mathrm{H} _{1}>\mathrm{H} _{3}>\mathrm{H} _{5}>$………………and $\mathrm{H} _{2}<\mathrm{H} _{4}<\mathrm{H} _{6}<$………………

(d) $\mathrm{H} _{1}<\mathrm{H} _{3}<\mathrm{H} _{5}<$………………and $\mathrm{H} _{2}>\mathrm{H} _{4}>\mathrm{H} _{6}>$………………

9. If $x, y, z>0$ and $x+y+z=1$, the $\dfrac{x y z}{(1-x)(1-y)(1-z)}$ is necessarily

(a) $\geq 8$

(b) $\leq \dfrac{1}{8}$

(c) $<\dfrac{1}{8}$

(d) None of these

10.* Match the following:

11. The coefficient of $\mathrm{x}^{203}$ is the expansion of $(\mathrm{x}-1)\left(\mathrm{x}^{2}-2\right)\left(\mathrm{x}^{2}-3\right)$……………..$\left(\mathrm{x}^{20}-20\right)$ is

(a) -35

(b) 21

(c) 13

(d) 25

12. Read the following passage and answer the questions :Let $\mathrm{ABCD}$ is a unit square and $0<\alpha<1$. Each side of the square is divided in the ratio $\alpha: 1-\alpha$, as shown in the figure. These points are connected to obtain another square. The sides of new square are divided in the ratio $\alpha: 1-\alpha$ and points are joined to obtain another square. The process is continued indefinitely.

Let $a _{n}$ denote the length of side and $A _{n}$ the area of the $n$th square

(i) The value of $\alpha$ for which $\sum \limits _{n=1}^{\infty} A _{n}=\dfrac{8}{3}$ is

(a) $1 / 3,2 / 3$

(b) $1 / 4,3 / 4$

(c) $1 / 5,4 / 5$

(d) $1 / 2$

(ii) The value of $\alpha$ for which side of $\mathrm{n}^{\text {th }}$ square equals the diagonals of $(\mathrm{n}+1)$ th square is

(a) $1 / 3$

(b) $1 / 4$

(c) $1 / 2$

(d) $1 / \sqrt{2}$

(iii) If $a=1 / 4$ and $P _{n}$ denotes the perimeter of the $n^{\text {th }}$ square then $\sum\limits _{n=1}^{\infty} P _{n}$ equals

(a) $8 / 3$

(b) $32 / 3$

(c) $16 / 3$

(d) $\dfrac{8}{3}(4+\sqrt{10})$