Logarithm

Laws of LogarithmSolved Examples on LogarithmCharacteristic and MantissaProperties of Logarithm Properties Of Monotonocity Of LogarithmLogarithmic Functions GraphLogarithm problems asked in Exams

Introduction to Logarithm

The logarithm of any positive number, whose base is a number, which is greater than zero and not equal to one, is the index or the power to which the base must be raised in order to obtain the given number.

Mathematically: If ax=b(wherea>0,1),{{a}^{x}}=b\left( where\,\,a>0,\ne 1 \right), then x is called the logarithm of b to the base a and we write loga b = x, clearly b > 0. Thus logab=xax=b,a>0,a1{{\log }_{a}}b=x\Leftrightarrow {{a}^{x}}=b,a>0,a\ne 1 and b > 0.

If a = 10, then we write log b rather than log10 b. If a = e, we write ln b rather than loge b. Here, ‘e’ is called as Napier’s base and has numerical value equal to 2.7182. Also, log10 e is known as Napierian constant.

i.e. log10 e = 0.4343

∴ ln b = 2.303 log10 b

Since, [lnb=log10b×loge10=1log10e×log10b=10.4343log10b=2303 log10b]\left[\ln \,b={{\log }_{10}}b\times {{\log }_{e}}10=\frac{1}{{{\log }_{10}}e}\times {{\log }_{10}}b \right.\left. =\frac{1}{0.4343}{{\log }_{10}}b=2303\ {{\log }_{10}}b \right]

⇒ Important Points

  1. log 2 = log10 2 = 0.3010
  2. log 3 = log10 3 = 0.4771
  3. ln 2 = 2.303 log 2 = 0.693
  4. ln 10 = 2.303

Laws of Logarithm

Corollary 1: From the definition of the logarithm of the number b to the base a, we have an identity

alogab=b,a>0,a1andb>0{{a}^{{{\log }_{a}}b}}=b,a>0,a\ne 1\,and\,b>0

Which is known as the Fundamental Logarithmic Identity.

Corollary 2: The function defined by f(x)=logax,a>0,a1f\left( x \right)={{\log }_{a}}x,a>0,a\ne 1 is called logarithmic function. In domain is (0,)\left( 0,\infty \right) and range is R (set of all real numbers).

Corollary 3: ax>0,xR{{a}^{x}}>0,\forall x\in R

  1. If a > 1, then ax is monotonically increasing. For example, 52.7>52.5,3222>3111{{5}^{2.7}}>{{5}^{2.5}},{{3}^{222}}>{{3}^{111}}
  2. If 0 < a < 1, then ax is monotonically decreasing. For example, (15)2.7<(15)2.5,(0.7)222<(0.7)212{{\left( \frac{1}{5} \right)}^{2.7}}<{{\left( \frac{1}{5} \right)}^{2.5}},{{\left( 0.7 \right)}^{222}}<{{\left( 0.7 \right)}^{212}}

Corollary 4:

  1. If a > 1, then a=0{{a}^{-\infty }}=0 i.e. loga0=8(ifa>1)\,\,\,\,\,{{\log }_{a}}0=-8\left( if\,a>1 \right)
  2. If 0 < a < 1, then u=0{{u}^{\infty }}=0 i.e. loga0=+(if0<a<1)\,\,\,\,\,{{\log }_{a}}0=+\infty \left( if\,0<a<1 \right)

Corollary 5:

  • logab,ifa>1,b{{\log }_{a}}b\to \infty ,if\,a>1,b\to \infty
  • logab,if0<a<1,b{{\log }_{a}}b\to -\infty ,if\,0<a<1,b\to \infty

Remarks

  1. ‘log’ is the abbreviation of the word ‘logarithm’.
  2. Common logarithm (Brigg’s logarithms) The base is 10.
  3. If x < 0, a > 0 and a1,a\ne 1, then loga x is an imaginary.
  4. loga1=0(a>0,a1){{\log }_{a}}1=0\left( a>0,a\ne 1 \right)
  5. logaa=1(a>0,a1){{\log }_{a}}a=1\left( a>0,a\ne 1 \right)
  6. log(1/a)a=1(a>0,a1){{\log }_{\left( 1/a \right)}}a=-1\left( a>0,a\ne 1 \right)

If a>1,logax={+ve,x>10,x=1ve,0<x<1a>1,{{\log }_{a}}x=\left\{ \begin{matrix} +ve,\,\,\,\,\,\,\,\,\,\,\,\,\,x>1 \\ 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x=1 \\ -ve,\,\,\,\,\,0<x<1 \\ \end{matrix} \right.

And if,

0<a<1,logax={+ve,0<x<10,x=1ve,x>10<a<1,{{\log }_{a}}x=\left\{ \begin{matrix} +ve,\,\,\,\,\,0<x<1 \\ 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x=1 \\ -ve,\,\,\,\,\,\,\,\,\,\,\,\,\,x>1 \\ \end{matrix} \right.

Solved Examples on Logarithm

Example: 1: Find the value of logtan45cot30{{\log }_{\tan 45{}^\circ }}\cot 30{}^\circ

Solution:

Here, base tan 450 = 1

∴ log is not defined.

Example: 2: Find the value of log(sec2600tan2600)cos600log_{(sec^2 60^0 – tan^2 60^0)} cos 60^0

Solution:

Here, base =sec260tan260=1={{\sec }^{2}}60{}^\circ -\tan^2 60{}^\circ =1

∴ log is not defined.

Example: 3: Find the value of log(sec230+cos230)1{{\log }_{\left( se{{c}^{2}}30{}^\circ +{{\cos }^{2}}30{}^\circ \right)}}1

Solution:

Since, log(sin230+cos230)1=log111\,\,{{\log }_{\left( {{\sin }^{2}}30{}^\circ +{{\cos }^{2}}30{}^\circ \right)}}1={{\log }_{1}}1\ne 1

Here, base = 1, ∴ log is not defined.

Example: 4: Find the value of log301{{\log }_{30}}1

Solution:

log301=0{{\log }_{30}}1=0

Characteristic and Mantissa

The integral part of a logarithm is called the characteristic and the fractional part (decimal part) is called mantissa.

i.e., log N = Integer + Fractional or decimal part (+ve)

⇒ The mantissa of the log of a number is always kept positive.

i.e., if log564 = 2.751279, then 2 is the characteristic and 0.751279 is the mantissa of the given number 564.

And if log0.00895 = -2.0481769 = -2–0.0481769 = (-2-1)+(1-0.0481769) = -3+0.9518231

Hence, -3 is the characteristic and 0.9518231

(not 0.0481769) is mantissa of log 0.00895.

In short, -3+0.951823 is written as 3.9518231.\overline{3}.9518231.

Important Conclusions on Characteristic And Mantissa

  • If the characteristics of log N be n, then the number of digits in N is (n+1) (Here, N > 1).
  • If the characteristics of log N be –n, then there exists (n-1) number of zeroes after decimal part of N (here, ) < N < 1).
  • If N > 1, the characteristic of log N will be on less than the number of digits in an integral part of N.
  • If 0 < N < 1, the characteristic of log N is negative and numerically it is one greater than the number of zeroes immediately after the decimal part in N.

For Example:

1. If log 235.68 = 2.3723227. Here, N = 235.68

∴ Number of digits in an integral part of N = 3

⇒ Characteristic of log 235.68 = N -1 = 3 – 1 = 2

2. If log  0.0000279=5.4456042\log\;0.0000279=\overline{5}.4456042

Here, four zeroes immediately after the decimal point in the number 0.0000279 is (4+1),i.e.5.\left( \overline{4+1} \right),i.e.\,\overline{5}.

Problem: If log 2 = 0.301 and log 3 = 0.477, find the number of digits in 620.

Solution: Let P = 620 = (2×3)20

∴ log P = 20 log (2×3) = 20 {log2+log3}

= 20 {0.301+0.477} = 20 × 0.778 = 15.560

Since, the characteristic of log P is 15, therefore the number of digits in P will be 15 + 1, i.e. 16.

Principle Properties of Logarithm

Let m and n be arbitrary positive numbers, be any real numbers, then

1. Loga (m n) = loga m + loga n

In general, loga (x1, x2, x3,…, xn) = loga x1

+logax2+logax3++logaxn+{{\log }_{a}}{{x}_{2}}+{{\log }_{a}}{{x}_{3}}+…+{{\log }_{a}}{{x}_{n}}

(where,x1,x2,x3,,xn>0)\left( where,\,{{x}_{1}},{{x}_{2}},{{x}_{3}},…,{{x}_{n}}>0 \right)

Or

loga(i=1nxi)=i=1nlogaxi,xi>0{{\log }_{a}}\left( \prod\limits_{i=1}^{n}{{{x}_{i}}} \right)=\sum\limits_{i=1}^{n}{{{\log }_{a}}{{x}_{i}},\forall {{x}_{i}}}>0

where, i = 1, 2, 3, …, n.

2. loga(mn)=logamlogan{{\log }_{a}}\left( \frac{m}{n} \right)={{\log }_{a}}m-{{\log }_{a}}n

3. logamα=αlogam{{\log }_{a}}{{m}^{\alpha }}=\alpha {{\log }_{a}}m

4. logaβm=1βlogam{{\log }_{{{a}^{\beta }}}}m=\frac{1}{\beta }{{\log }_{a}}m

5. logbm=logamlogab{{\log }_{b}}m=\frac{{{\log }_{a}}m}{{{\log }_{a}}b}

6. logba.logab=1logba=1logab{{\log }_{b}}a.{{\log }_{a}}b=1\,\,\,\,\,\Leftrightarrow \,\,\,\,\,{{\log }_{b}}a=\frac{1}{{{\log }_{a}}b}

7. logba.logcb.logac=1{{\log }_{b}}a.{{\log }_{c}}b.{{\log }_{a}}c=1

8. logyx.logzy.logaz=logax{{\log }_{y}}x.{{\log }_{z}}y.{{\log }_{a}}z={{\log }_{a}}x

9. elnax=ax{{e}^{\ln \,{{a}^{x}}}}={{a}^{x}}

Some Additional Logarithm Properties

  1. alogbx=xlogba,b1,a,b,x{{a}^{{{\log }_{b}}x}}={{x}^{{{\log }_{b}}a}},b\ne 1,a,b,x are positive numbers.
  2. alogax=x,a>0,a1,x>0{{a}^{{{\log }_{a}}x}}=x,a>0,a\ne 1,x>0
  3. logakx=1klogax,a>0,a1,x>0{{\log }_{{{a}^{k}}}}x=\frac{1}{k}{{\log }_{a}}x,a>0,a\ne 1,x>0
  4. logax2k=2klogax,a>0,a1,kI{{\log }_{a}}{{x}^{2k}}=2k{{\log }_{a}}\left| x \right|,a>0,a\ne 1,k\in I
  5. loga2kx=12klogax,x>0,a±1andkI ~{10}{{\log }_{{{a}^{^{2k}}}}}x=\frac{1}{2k}{{\log }_{\left| a \right|}}x,x>0,a\ne \pm 1\,\,and\,\,k\in I\tilde{\ }\left\{ 10 \right\}
  6. logaαxβ=βαlogax,x>0,a>0,a1,α0{{\log }_{{{a}^{\alpha }}}}{{x}^{\beta }}=\frac{\beta }{\alpha }{{\log }_{a}}x,x>0,a>0,a\ne 1,\alpha \ne 0
  7. logax22logax,a>0,a1{{\log }_{a}}{{x}^{2}}\ne 2{{\log }_{a}}x,a>0,a\ne 1

Since, domain of loga(x)2{{\log }_{a}}{{\left( x \right)}^{2}} is R ~ {0} and domain of logaxis(0,){{\log }_{a}}x\,\,is\,\,\left( 0,\infty \right) are not same.

  1. ablog  a=a,ifb=a2,a>0,b>0,b1a_{b}^{log\;a} =\sqrt{a},\,if\,b={a}^{2},a>0,b>0,b\ne 1
  2. ablog  a=a2,ifb=a,a>0,b>0,b1a_{b}^{log\;a} =a^{2},\,if\,b=\sqrt{a},a>0,b>0,b\ne 1

Example 1: Solve the equation 3.xlog5  2+2log5  x=643.x^{\log_{5}\;2} + 2^{\log_{5}\;x}=64

Solution:

3.xlog5  2+2log5  x=643.x^{\log_{5}\;2} + 2^{\log_{5}\;x}=64

3.2log5  x+2log5  x=643.2^{\log_{5}\;x} + 2^{\log_{5}\;x}=64 [by extra property (i)]

4.2log5x=644.2^{\log_5x} =64

2log5x=42=242^{\log _{5}x}=4^2=2^4

∴ log5x=4\log _{5}x=4

x=54=625x=5^4=625

Example 2: If 4log164+9log39=10logx83,{{4}^{{{\log }_{16\,}}4}}+{{9}^{{{\log }_{3}}9}}={{10}^{{{\log }_{x}}83}}, find x.

Solution:

Since, 4log164=4=2{{4}^{{{\log }_{16}}4}}=\sqrt{4}=2 [by extra property (ix)]

and 9log39=92=81{{9}^{{{\log }_{3}}9}}={{9}^{2}}=81 [by extra property (viii)]

4log164+9log39=2+81=83=10logx83{{4}^{{{\log }_{16}}4}}+{{9}^{{{\log }_{3}}9}}=2+81=83={{10}^{{{\log }_{x}}83}}

log1083=logx83{{\log }_{10}}83={{\log }_{x}}83

∴ x = 10

Example 3: Prove that alogabblogba=0.{{a}^{\sqrt{{{\log }_{a}}b}}}-{{b}^{\sqrt{{{\log }_{b}}a}}}=0.

Solution:

Since, a(logab)=alogab×logab×logba\,\,\,\,\,{{a}^{\sqrt{\left( {{\log }_{a}}b \right)}}}={{a}^{\sqrt{{{\log }_{a}}b}\,\times \,\sqrt{{{\log }_{a}}b}\,\times \,\sqrt{{{\log }_{b}}a}}}

= alogab.logba{{a}^{{{\log }_{a}}b\,.\,\,\sqrt{{{\log }_{b}}a}}}

= blogba{{b}^{\sqrt{{{\log }_{b}}a}}} [by extra property (ii)]

Hence, a(logab)b(logba)=0{{a}^{\sqrt{\left( {{\log }_{a}}b \right)}}}-{{b}^{\sqrt{\left( {{\log }_{b}}a \right)}}}=0

Example 4: Prove that log224log962log2192log122=3.\frac{{{\log }_{2}}24}{{{\log }_{96}}2}-\frac{{{\log }_{2}}192}{{{\log }_{12}}2}=3.

Solution:

LHS=log224log962log2192log122LHS=\frac{{{\log }_{2}}24}{{{\log }_{96}}2}-\frac{{{\log }_{2}}192}{{{\log }_{12}}2}

= log224×log296log2192×log212{{\log }_{2}}24\times {{\log }_{2}}96-{{\log }_{2}}192\times {{\log }_{2}}12

Now, let 12=λ,12=\lambda , then

LHS=log22λ×log28λlog216λ×log2λLHS={{\log }_{2}}2\lambda \times {{\log }_{2}}8\lambda -{{\log }_{2}}16\lambda \times {{\log }_{2}}\lambda

= (log22+log2λ)(log28+log2λ)\left( {{\log }_{2}}2+{{\log }_{2}}\lambda \right)\left( {{\log }_{2}}8+{{\log }_{2}}\lambda \right)

(log216+log2λ)log2λ-\left( {{\log }_{2}}16+{{\log }_{2}}\lambda \right){{\log }_{2}}\lambda

= (log22+log2λ)(log223+log2λ)\left( {{\log }_{2}}2+{{\log }_{2}}\lambda \right)\left( {{\log }_{2}}{{2}^{3}}+{{\log }_{2}}\lambda \right)

(log224+log2λ)log2λ-\left( {{\log }_{2}}{{2}^{4}}+{{\log }_{2}}\lambda \right){{\log }_{2}}\lambda

= (1+log2λ)(3log22+log2λ)\left( 1+{{\log }_{2}}\lambda \right)\left( 3{{\log }_{2}}2+{{\log }_{2}}\lambda \right)

(4log22+log2λ)log2λ-\left( 4{{\log }_{2}}2+{{\log }_{2}}\lambda \right){{\log }_{2}}\lambda

= (1+log2λ)(3+log2λ)log2λ(4+log2λ)\left( 1+{{\log }_{2}}\lambda \right)\left( 3+{{\log }_{2}}\lambda \right)-{{\log }_{2}}\lambda \left( 4+{{\log }_{2}}\lambda \right)

= 3 = RHS

Properties Of Monotonocity Of Logarithm

Logarithm with Constant Base

  1. logax>logay{x>y>0,ifa>10<x<y,if0<a<1{{\log }_{a}}x>{{\log }_{a}}y\Leftrightarrow \left\{ \begin{matrix} x>y>0,\,if\,a>1 \\ 0<x<y,\,if\,0<a<1 \\ \end{matrix} \right.
  2. logax<logay{0<x<y,ifa>1x>y>0,if0<a<1{{\log }_{a}}x<{{\log }_{a}}y\Leftrightarrow \left\{ \begin{matrix} 0<x<y,\,if\,a>1 \\ x>y>0,\,if\,0<a<1 \\ \end{matrix} \right.
  3. logax>p{x>ap,ifa>10<x<ap,if0<a<1{{\log }_{a}}x>p\Leftrightarrow \left\{ \begin{matrix} x>{{a}^{p}},\,if\,a>1 \\ 0<x<{{a}^{p}},\,if\,0<a<1 \\ \end{matrix} \right.
  4. logax<p{0<x<ap,ifa>1x>ap,if0<a<1{{\log }_{a}}x<p\Leftrightarrow \left\{ \begin{matrix} 0<x<{{a}^{p}},\,if\,a>1 \\ x>{{a}^{p}},\,if\,0<a<1 \\ \end{matrix} \right.

Logarithm with Variable Base

  1. logx a is defined, if a>0,x>0,x1a>0,x>0,x\ne 1
  2. If a > 1, then logx a is monotonically decreasing in (0,1)(1,)\left( 0,1 \right)\cup \left( 1,\infty \right)
  3. If 0 < a < 1, then logx a is monotonically increasing in (0,1)(1,)\left( 0,1 \right)\cup \left( 1,\infty \right)

Key Points

  1. If a > 1, p > 1, then loga p > 0
  2. If 0 < a < 1, p > 1, then loga p < 0
  3. If a > 1, 0 < p < 1, then loga p < 0
  4. If p > a > 1, then loga p > 1
  5. If a > p > 1, then 0 < loga p < 1
  6. If 0 < a < p < 1, then 0 < loga p < 1
  7. If 0 < p < a < 1, then loga p > 1

Graphs Of Logarithmic Functions

1. Graph of y = loga x, if a > 1 and x > 0

2. Graph of y = loga x, if 0 < a < 1 and x > 0

If the number x and the base ‘a’ are on the same side of the unity, then the logarithm is positive.

  • y = loga x, a > 1, x > 1
  • y = loga x, 0 < a < 1, 0 < x < 1

If the number x and the base a are on the opposite sides of the unity, then the logarithm is negative.

  • y = loga x, a > 1, 0 < x < 1
  • y = loga x, 0 < a < 1, x > 1

3. Graph of y=logaxy={{\log }_{a}}\left| x \right|

Graphs are symmetrical about Y-axis.

4. Graph of y=logaxy=\left| {{\log }_{a}} \right|\left. x \right\|

Graphs are same in both cases i.e., a > 1 and 0 < a < 1.

5. Graph of y=logax\left| y \right|=\left. {{\log }_{a}} \right|\left. x \right\|

6. Graph of y=loga[x],a>1andx1y={{\log }_{a}}\left[ x \right],a>1\,and\,x\ge 1

(where [ . ] denotes the greatest integer function)

Since, when 1x<2,[x]=1loga[x]=01\le x<2,\left[ x \right]=1\Rightarrow {{\log }_{a}}\left[ x \right]=0

when 2x<3,[x]=2loga[x]=loga22\le x<3,\left[ x \right]=2\Rightarrow {{\log }_{a}}\left[ x \right]={{\log }_{a}}2

when 3x<4,[x]=3loga[x]=loga33\le x<4,\left[ x \right]=3\Rightarrow {{\log }_{a}}\left[ x \right]={{\log }_{a}}3 so on.

Important Shortcuts to Answer Logarithm Problems

  1. For a non-negative number ‘a’ and n2,nN,an=a1/n.n\ge 2,n\in N,\sqrt[n]{a}={{a}^{1/n}}.
  2. The number of positive integers having base a and characteristic n is an+1an.{{a}^{n+1}}-{{a}^{n}}.
  3. Logarithm of zero and negative real number is not defined.
  4. logba+logab2,a>0,a1,b>0,b1\left| {{\log }_{b}}a+{{\log }_{a}}b \right|\ge 2,\forall a>0,a\ne 1,b>0,b\ne 1
  5. log2log22ntimes=n{{\log }_{2}}{{\log }_{2}}\underbrace{\sqrt{\sqrt{\sqrt{\sqrt{…\sqrt{2}}}}}}_{n\,\,times}=-n
  6. alogab=blogba{{a}^{\sqrt{{{\log }_{a}}b}}}={{b}^{\sqrt{{{\log }_{b}}a}}}
  7. Logarithms to the base 10 are called common logarithms (Brigg’s logarithms).
  8. If x=logcb+logbc,y=logac+logca,z=logab+logba,thenx2+y2+z24=xyz.x={{\log }_{c}}b+{{\log }_{b}}c,y={{\log }_{a}}c+{{\log }_{c}}a,z={{\log }_{a}}b+{{\log }_{b}}a,\,\,then\,\,{{x}^{2}}+{{y}^{2}}+{{z}^{2}}-4=xyz.

Practice Problems on Logarithm

Problem 1: If log 11 = 1.0414, prove that 1011 > 1110.

Solution:

log1011=11log10=11\log {{10}^{11}}=11\log 10=11

and log1110 = 10log11 = 10 × 1.0414 = 10.414

It is clear that, 11 > 10.414

log1011>log1110\log {{10}^{11}}>\log {{11}^{10}} [Since, base = 10]

1011>1110{{10}^{11}}>{{11}^{10}}

Problem 2: If log2 (x-2)<log4 (x-2), find the interval in which x lies.

Solution:

Here, x – 2 > 0

⇒ x > 2 ……………… (i)

and log2(x2)<log22(x2)=12log2(x2){{\log }_{2}}\left( x-2 \right)<{{\log }_{{{2}^{2}}}}\left( x-2 \right)=\frac{1}{2}{{\log }_{2}}\left( x-2 \right)

log2(x2)<12log2(x2){{\log }_{2}}\left( x-2 \right)<\frac{1}{2}{{\log }_{2}}\left( x-2 \right)

12log2(x2)<0log2(x2)<0\frac{1}{2}{{\log }_{2}}\left( x-2 \right)<0\Rightarrow {{\log }_{2}}\left( x-2 \right)<0

x2<20x2<1 x-2<{{2}^{0}}x-2<1

⇒ x < 3 ……………… (ii)

From equations (i) and (ii), we get

2<x<3orx(2,3)2<x<3\,\,or\,\,x\in \left( 2,3 \right)

Problem 3: If alogbc=3.3log43.3log43log43.3log43log43logbc{{a}^{{{\log }_{b}}c}}={{3.3}^{{{\log }_{4}}3}}{{.3}^{{{\log }_{4}}3}}^{^{{{\log }_{4}}3}}{{.3}^{{{\log }_{4}}{{3}^{^{{{\log }_{4}}{{3}^{{{\log }_{b}}c}}}}}}}…\infty

where, a,b,cQ,a,b,c\in Q, the value of abc is

(a) 9 (b) 12 (c) 16 (d) 20

Solution:

(c) alogbc=31+log43+(log43)2+(log43)3+{{a}^{{{\log }_{b}}c}}={{3}^{1+{{\log }_{4}}3}}{{+}^{{{\left( {{\log }_{4}}3 \right)}^{2}}}}{{+}^{{{\left( {{\log }_{4}}3 \right)}^{3}}}}+…\infty

= 31/(1log43)=31/log4(4/3)=3log4/34{{3}^{1/\left( 1-{{\log }_{4}}3 \right)}}={{3}^{1/{{\log }_{4}}\left( 4/3 \right)}}={{3}^{{{\log }_{4/3}}4}}

a=3,b=43,c=4a=3,b=\frac{4}{3},c=4

Hence, abc=3.43.4=16abc=3.\frac{4}{3}.4=16

Problem 4: Number of real roots of equation 3log3(x24x+3)=(x3){{3}^{{{\log }_{3}}\left( {{x}^{2}}-4x+3 \right)}}=\left( x-3 \right) is

(a) 0 (b) 1 (c) 2 (d) infinite

Solution:

3log3(x24x+3)=(x3){{3}^{{{\log }_{3}}\left( {{x}^{2}}-4x+3 \right)}}=\left( x-3 \right) ……….(i)

Equation (i) is defined, if x2 – 4x + 3 > 0

(x1)(x3)>0\,\,\left( x-1 \right)\left( x-3 \right)>0

⇒ x < 1 or x > 3 ……….(ii)

Equation (i) reduces to x24x+3=x3x25x+6=0{{x}^{2}}-4x+3=x-3\Rightarrow {{x}^{2}}-5x+6=0

∴ x = 2, 3 ……….(iii)

From equations (ii) and (iii), use get xϕx\in \phi

∴ Number of real roots = 0.

Problem 5:  If log6 a+ log6 b+ log6 c = 6, where a,b,cNa,b,c\in N and a, b, c are in GP and b – a is a square of an integer, then the value of a + b – c is

(a) 21 (b) 15 (c) 9 (d) 3

Solution:

(b) Since, log6 a+ log6 b+ log6 c = 6

log6(abc)=6{{\log }_{6}}\left( abc \right)=6

∴ abc =6

b3=66[Since,b2=ac]{{b}^{3}}\,={{6}^{6}}\,\,\,\,\,\,\left[ Since, {{b}^{2}}=ac \right]

⇒ b = 36

Problem 6:  If x=log2a(bcd2),y=log3b(acd3),z=log4c(abd4)x={{\log }_{2a}}\left( \frac{bcd}{2} \right),y={{\log }_{3b}}\left( \frac{acd}{3} \right),z={{\log }_{4c}}\left( \frac{abd}{4} \right) and w=log5d(abc5)w={{\log }_{5d}}\left( \frac{abc}{5} \right) and 1x+1+1y+1+1z+1+1w+1=logabcdN+1,\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}+\frac{1}{w+1}={{\log }_{abcd}}N+1, the value of N is

(a) 40 (b) 80 (c) 120 (d) 160

Solution:

(c)

Since, x=log2a(bcd2)x={{\log }_{2a}}\left( \frac{bcd}{2} \right)

x+1=log2a(2abcd2)=log2a(abcd)x+1={{\log }_{2a}}\left( \frac{2abcd}{2} \right)={{\log }_{2a}}\left( abcd \right)

1x+1=logabcd2a\frac{1}{x+1}={{\log }_{abcd}}2a

Similarly, 1y+1=logabcd3b,1z+1=logabcd4c\frac{1}{y+1}={{\log }_{abcd}}3b,\frac{1}{z+1}={{\log }_{abcd}}4c

and 1w+1=logabcd5d\frac{1}{w+1}={{\log }_{abcd}}5d

1x+1+1y+1+1z+1+1w+1=logabcd(2a3b4c5d)\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}+\frac{1}{w+1}={{\log }_{abcd}}\left( 2a\cdot 3b\cdot 4c\cdot 5d \right)

= logabcd(120abcd){{\log }_{abcd}}\left( 120abcd \right)

= logabcd120+1{{\log }_{abcd}}120+1

= logabcdN+1{{\log }_{abcd}}N+1 [given]

Hence, N = 120

Problem 7:  If a = log12 18, b = log24 54 then the value of ab + (a–b) is

(a) 0 (b) 4 (c) 1 (d) none of these

Solution:

(c) we have

a=log1218=log218log212=1+2log232+log23a={{\log }_{12}}18=\frac{{{\log }_{2}}18}{{{\log }_{2}}12}=\frac{1+2{{\log }_{2}}3}{2+{{\log }_{2}}3} and

b=log2454=log254log224=1+3log233+log23b={{\log }_{24}}54=\frac{{{\log }_{2}}54}{{{\log }_{2}}24}=\frac{1+3{{\log }_{2}}3}{3+{{\log }_{2}}3}

Putting x = log2 3, we have

ab+5(ab)=1+2x2+x1+3x3+x+5(1+2x2+x1+3x3+x)ab+5\left( a-b \right)=\frac{1+2x}{2+x}\cdot \frac{1+3x}{3+x}+5\left( \frac{1+2x}{2+x}-\frac{1+3x}{3+x} \right)

=6x2+5x+1+5(x2+1)(x+2)(x+3)=x2+5x+6(x+2)(x+3)=1=\frac{6{{x}^{2}}+5x+1+5\left( -{{x}^{2}}+1 \right)}{\left( x+2 \right)\left( x+3 \right)}=\frac{{{x}^{2}}+5x+6}{\left( x+2 \right)\left( x+3 \right)}=1.

Problem 8: The value of log224log962log2192log122\frac{{{\log }_{2}}24}{{{\log }_{96}}2}-\frac{{{\log }_{2}}192}{{{\log }_{12}}2} is

(a) 3 (b) 0 (c) 2 (d) 1

Solution:

(a) Set log2 12 = a,

1log962=log296=log223×12=3+a,\frac{1}{{{\log }_{96}}2}={{\log }_{2}}96={{\log }_{2}}{{2}^{3}}\times 12=3+a,

log224=1+a,log2192=log2(16×12)=4+a{{\log }_{2}}24=1+a,{{\log }_{2}}192={{\log }_{2}}\left( 16\times 12 \right)=4+a

and 1log122=log212=a.\frac{1}{{{\log }_{12}}2}={{\log }_{2}}12=a.

Therefore, the given expression

=(1+a)(3+a)(4+a)a=3=\left( 1+a \right)\left( 3+a \right)-\left( 4+a \right)a=3

Problem 9:  The solution of the equation 4log2logx=logx(logx)2+1{{4}^{{{\log }_{2}}\log x}}=\log x-{{\left( \log x \right)}^{2}}+1 is

(a) x = 1 (b) x = 4 (c) x = 3 (d) x = e2

Solution:

(c) log2 log x is meaningful if x > 1

Since 4log2logx=22log2logx=(2log2logx)2=(logx)2{{4}^{{{\log }_{2}}\log x}}={{2}^{2{{\log }_{2}}\log x}}={{\left( {{2}^{{{\log }_{2}}\log x}} \right)}^{2}}={{\left( \log x \right)}^{2}}

[alogax=x,a>0,a1]\left[ {{a}^{{{\log }_{a}}x}}=x,a>0,a\ne 1 \right]

So the given equation reduces to

2(logx)2logx1=02{{\left( \log x \right)}^{2}}-\log x-1=0

logx=1,logx=1/2.\log x=1,\log x=-1/2. But for x > 1

log x > 0 so log x = 1 i.e. x = 3.

Problem 10:  If log x2 – log 2 x = 3 log 3 – log 6 then x equals

(a) 9 (b) 3 (c) 4 (d) 5

Solution:

(a) Clearly x > 0. The given equation can be written as

2 log x – log 2 – log x = 3 log 3 – log 2 – log 3

logx=2,log3x=9.\log x=2,\log 3\Rightarrow x=9.

Problem 11:  If log0.5 (x – 1) < log0.25 (x – 1), then x lies in the interval.

  • (2,)\left( 2,\infty \right)
  • (3,)\left( 3,\infty \right)
  • (,0)\left( -\infty ,0 \right) (d) (0, 3)

Solution:

log0.5 (x – 1) < log0.25 (x – 1)

log0.5(x1)<log(0.5)2(x1)\Leftrightarrow {{\log }_{0.5}}\left( x-1 \right)<{{\log }_{{{\left( 0.5 \right)}^{2}}}}\left( x-1 \right)

log0.5(x1)<log0.5(x1)log0.5(0.5)2=12log0.5(x1)\Rightarrow {{\log }_{0.5}}\left( x-1 \right)<\frac{{{\log }_{0.5}}\left( x-1 \right)}{{{\log }_{0.5}}{{\left( 0.5 \right)}^{2}}}=\frac{1}{2}{{\log }_{0.5}}\left( x-1 \right)

log0.5(x1)<0\Leftrightarrow \,\,\,\,{{\log }_{0.5}}\left( x-1 \right)<0

x1>1x(2,)\Leftrightarrow \,\,\,\,\,\,\,x-1>1\Leftrightarrow x\in \left( 2,\infty \right)

Problem 12:  If n = 2002 !, evaluate 1log2n+1log3n+1log4n++1log2002n\frac{1}{{{\log }_{2}}n}+\frac{1}{{{\log }_{3}}n}+\frac{1}{{{\log }_{4}}n}+…+\frac{1}{{{\log }_{2002}}n}

Solution:

We have, 1log2n+1log3n+1log4n++1log2002n\frac{1}{{{\log }_{2}}n}+\frac{1}{{{\log }_{3}}n}+\frac{1}{{{\log }_{4}}n}+…+\frac{1}{{{\log }_{2002}}n}

=logn2+logn3+logn4++logn2002 ={{\log }_{n}}2+{{\log }_{n}}3+{{\log }_{n}}4+…+{{\log }_{n}}2002  Since, 1logba=logab\frac{1}{{{\log }_{b}}a}={{\log }_{a}}b

= logn(2.3.4.2002){{\log }_{n}}\left( 2.3.4….2002 \right)

= logn(2002!)=lognn=1{{\log }_{n}}\left( 2002! \right)={{\log }_{n}}n=1

Problem 13:  If x, y, z > 0 and such that logxyz=logyzx=logzxy\frac{\log x}{y-z}=\frac{\log y}{z-x}=\frac{\log z}{x-y} prove that xx yy zz = 1

Solution:

Let logxyz=logyzx=logzxy=λ\frac{\log x}{y-z}=\frac{\log y}{z-x}=\frac{\log z}{x-y}=\lambda

logx=λ(yz),logy=λ(zx),logz=λ(xy)\log x=\lambda \left( y-z \right),\log y=\lambda \left( z-x \right),\log z=’\lambda \left( x-y \right)

xlogx+ylogy+zlogzx\log x+y\log y+z\log z

=λx(yz)+λy(zx)+λz(xy)=0=\lambda x\left( y-z \right)+\lambda y\left( z-x \right)+\lambda z\left( x-y \right)=0

logxx+logyy+logzz=0\log {{x}^{x}}+\log {{y}^{y}}+\log {{z}^{z}}=0

log(xxyyzz)=0\log \left( {{x}^{x}}{{y}^{y}}{{z}^{z}} \right)=0

xxyyzz=1{{x}^{x}}{{y}^{y}}{{z}^{z}}=1

Problem 14:  Solve: log3 {5 + 4 log3 (x – 1)} = 2

Solution:

Clearly, the given equation is meaningful, if x – 1 > 0 and 5 + 4 log3 (x – 1) > 0

x>1andlog3(x1)>54x>1\,and\,{{\log }_{3}}\left( x-1 \right)>-\frac{5}{4}

x>1andx1>35/4x>1\,and\,x-1>{{3}^{-5/4}}

x>1andx>1+135/4x>1\,and\,x>1+\frac{1}{{{3}^{5/4}}}

x>1+135/4x>1+\frac{1}{{{3}^{5/4}}} ………. (i)

Now,

log3 {5 + 4 log3 (x – 1)} = 2

5+4log3(x1)=325+4{{\log }_{3}}\left( x-1 \right)={{3}^{2}}

log3(x1)=1{{\log }_{3}}\left( x-1 \right)=1

⇒ x – 1 = 3

∴ x=4

Clearly, x = 4 satisfies (i).

Hence, x = 4 is the solution to the given equation.

Problem 15: Solve log3 (3x – 8) = 2 – x.

Solution: Clearly, the given equation is meaningful, if3x8>03x>8x>log38{{3}^{x}}-8>0\Rightarrow {{3}^{x}}>8\Rightarrow x>{{\log }_{3}}8 ………. (i)

Now,

log3 (3x – 8) = 2 – x

(3x8)=2x\left( {{3}^{x}}-8 \right)=2-x

3x8=323x{{3}^{x}}-8=\frac{{{3}^{2}}}{{{3}^{x}}}

(3x)28(3x)9=0{{\left( {{3}^{x}} \right)}^{2}}-8\left( {{3}^{x}} \right)-9=0