Chapter – 21 : Logarithm | Chapter Solution Class 9

(iii) \log _{\sqrt{6}}^{216}

(iv) \log _{4}^{(\frac{1}{6-1})}

Solution:

(i) \log _{2 \sqrt{3}}^{1728}

=\log _{2}^{12^{3}}

=\log _{2 \sqrt{3}}^{(2 \sqrt{3})^{2 \times 3}}

=\log _{2 \sqrt{3}}^{(2 \sqrt{3})^6}

=6 \log _{2 \sqrt{3}}^{2 \sqrt{3}}

=6 \times 1=6 \text { (Ans) }

(ii) \log _{0.01}^{0.000001}

=\log _{10^{-2}}^{10^{-6}}

=\frac{-6}{-2} \log _{10}^{10}

=3 \times 1=3(\text { Ans })

(iii) \log _{\sqrt{6}}^{216}

=\log _{\frac{1}{6^{2}}}^{6^{3}}

=\frac{3}{\frac{1}{2}} \log _{6}^{6}

=3 \times 2 \log _{6}^{6}

=6 \times 1

=6(\text { Ans })

(iv) \log _{4}^{(\frac{1}{64})}

=\log _{2^{2}}^{(\frac{1}{2^{6}})}

=\log _{2^{2}}^{2^{-6}}

=\frac{-6}{2} \log _{2}^{2}

=-3 \times 1

=-3(\text { Ans })

2. (a) let us write by calculating, find its base if logarithm of 625 is 4

(b) let us write by calculating, find its base if logarithm 5832 is 6

Solution:

(a) Let the base be x

\log _{x}^{625}=4

or, x^{4}=625

or, x^{4}=5^{4}

\therefore x = 5

\therefore The base = 5( Ans )

(b) Let the base be x

\log _{x}^{5832}=6

or, x^{6}=5832

or, x^{6}=(3 \sqrt{2})^{6}

\therefore x = 3 \sqrt{2}

\therefore The base =3 \sqrt{2} (Ans)

3. (a) If 1+\log _{10}^{\mathrm{a}}=2 \log _{10}^{\mathrm{b}}, then express a by b

(b) If 3+\log _{10}^{x}=2 \log _{10}^{y}, then express x by y

Solution:

(a) 1+\log _{10}^{a}=2 \log _{10}^{b}

or, \log _{10}^{10}+\log _{10}^{a}=\log _{10}^{b^{2}} \quad[\because \log _{10}^{10}=1]

or, \log _{10}^{10 a}=\log _{10}^{b^{2}}

or, 10 a=b^{2}

\therefore, a=\frac{b^{2}}{10} (Ans)

(b) 3+\log _{10}^{1}=2 \cdot \log _{10}^{y}

or, 3 \log _{10}^{10}+\log _{10}^{x}=\log _{10}^{y^{2}}[\because \log _{10}^{10}=1]

or, \log _{10}^{10^{3}}+\log _{10}^{x}=\log _{10}^{y^{2}}

or, \log _{10}^{10^{3} x}=\log _{10}^{y^{2}}

or, 10^{3} x=y^{2}

or, 1000 x=y^{2} \therefore x=\frac{y^{2}}{1000}(Ans)

4. Let us evaluate:

(a) \log _{2}[\log _{2}\{\log _{3}(\log _{3}^{27^{3}})\}]

Solution:

\log _{2}[\log _{2}\{\log _{3}(\log _{3}^{27^{3}})\}]

= \log _{2}[\log _{2}\{\log _{3}(3 \log _{3}^{27})\}]

= \log _{2}[\log _{2}\{\log _{3}(3 \log _{3}^{3^{3}})]

= \log _{2}[\log _{2}\{\log _{3}(9 \log _{3}^{3})\}]

= \log _{2}[\log _{2}\{\log _{3}(9)\}]

= \log _{2}[\log _{2}\{\log _{3}^{9}\}]

= \log _{2}[\log _{2}\{\log _{3}^{3^{3}}\}]

= \log _{2}[\log _{2}\{2 \log _{3}^{3}\}]

= \log _{2}[\log _{2}\{2\}]

= \log _{2}[\log _{2}^{2}]

= \log _{2}[1][\therefore \log _{2}^{2}=1]

= \log _{2}^{1}

= 0 (Ans)

(b) \frac{\log ^{\sqrt{27}}+\log ^{8}-\log ^{\sqrt{1000}}}{\log ^{1.2}}

Solution:

\frac{\log ^{\sqrt{27}}+\log ^{8}-\log ^{\sqrt{1000}}}{\log ^{1.2}}

=\frac{\log ^{\sqrt{3^{3}}}+\log ^{2^{3}}-\log ^{\sqrt{10^{3}}}}{\log ^{\frac{12}{10}}}

=\frac{\log ^{\frac{3}{2}}+3 \log ^{2}-\log ^{10^{\frac{3}{2}}}}{\log ^{12}-\log ^{10}}

=\frac{\frac{3}{2} \log ^{3}+3 \log ^{2}-\frac{3}{2} \log ^{10}}{\log ^{2^{2} .3}-1}

=\frac{\frac{3}{2} \log ^{3}+3 \log ^{2}-\frac{3}{2} \times 1}{\log ^{2^{2}}+\log ^{3}-1}

=\frac{3 \log ^{2}+\frac{3}{2} \log ^{3}-\frac{3}{2}}{2 \log ^{2}+\log ^{3}-1}

=\frac{\frac{3}{2}(2 \log ^{2}+\log ^{3}-1)}{(2 \log ^{2}+\log ^{3}-1)}=\frac{3}{2}(\text { Ans })

(c) \log _{3}^{4} \times \log _{4}^{5} \times \log _{5}^{6} \times \log _{6}^{7} \times \log _{7}^{3}

Solution:

\log _{3}^{4} \times \log _{4}^{5} \times \log _{5}^{6} \times \log _{6}^{7} \times \log _{7}^{3}

= \frac{\log 4}{\log 3} \times \frac{\log 5}{\log 4} \times \frac{\log 6}{\log 5} \times \frac{\log 7}{\log 6} \times \frac{\log 3}{\log 7}

= 1 (Ans)

(d) \log _{10}^{\frac{384}{5}}+\log _{10}^{\frac{81}{32}}+3 \log _{10}^{\frac{5}{3}}+\log _{10}^{\frac{1}{9}}

Solution:

\log _{10}^{\frac{384}{5}}+\log _{10}^{\frac{81}{32}}+3 \log _{10}^{\frac{5}{3}}+\log _{10}^{\frac{1}{9}}

= \log _{10}^{384}-\log _{10}^{5}+\log _{10}^{81}-\log _{10}^{32}+3 \log _{10}^{5}-3 \log _{10}^{3}+\log _{100}^{1}+\log _{10}^{9}

= \log _{10}^{2^{7} .3}+\log _{10}^{3^{4}}-\log _{10}^{2^{5}}+2 \log _{10}^{5}-3 \log _{10}^{3}-\log _{10}^{3^{2}}[\therefore \log _{10}^{1}=0]

= \log _{10}^{2^{7}}+\log _{10}^{3}+4 \log _{10}^{3}-5 \log _{10}^{2}+2 \log _{10}^{5}-3 \log _{10}^{3}-2 \log _{10}^{3}

= 7 \log _{10}^{2}+5 \log _{10}^{3}-5 \log _{10}^{2}+2 \log _{10}^{5}-5 \log _{10}^{3}

= 2 \log _{10}^{2}+2 \log _{10}^{5}

= 2(\log _{10}^{2}+\log _{10}^{5})

= 2 . \log _{10}^{25.5}

= 2 \log _{10}^{10}

= 2 \times 1=2(\text { Ans })

5. let us prove :

(i) \lg \frac{75}{16}-2 \log \frac{5}{9}+\log \frac{32}{243}=\log2

Solution:

\text { LHS }=\log \frac{75}{16}-2 \log \frac{5}{9}+\log \frac{32}{243}

=\log 75-\log 16-2(\log 5-\log 9)+\log 32-\log 243

=\log 5^{2} \cdot 3-\log 4^{4}-2 \log 5+2 \log 3^{2}+\log 2^{5}-\log 3^{5}

=\log 5^{2}+\log 3-4 \log 2-2 \log 5+2 \times 2 \log 3+5 \log 2-5 \log 3

=2 \log 5+\log 3-4 \log 2-2 \log 5+4 \log 3+5 \log 2-5 \log 3

=5 \log 3+\log 2-5 \log 3

=\log 2

\therefore LHS = RHS (proved)

(ii) \log _{00}^{5}(1+\log _{15}^{30})+\frac{1}{2} \log _{00}^{16}(1+\log _{4}^{7})+\log _{0}^{6}(\log _{6}^{3}+1+\log _{6}^{7})=2

Solution:

\text { LHS }=\log _{10}^{15}(1+\log _{15}^{30})+\frac{1}{2} \log _{10}^{16}(1+\log _{4}^{7})-\log _{10}^{6}(\log _{6}^{3}+1+\log _{6}^{7})

=\log _{10}^{15}+\log _{10}^{15} \cdot \log _{15}^{30}+\frac{1}{2} \log _{10}^{16}+\frac{1}{2} \log _{10}^{16} \cdot \log _{7}^{4}-\log _{10}^{6} \cdot \log _{3}^{6}-\log _{10}^{6}-\log _{10}^{6} \cdot \log _{6}^{7}

=\log _{10}^{53}+\frac{\log 15}{\log 10} \times \frac{\log 30}{\log 15}+\frac{1}{2} \log _{10}^{2^{4}}+\frac{1}{2} \log _{10}^{2^{4}} \cdot \log _{2^{2}}^{7} \frac{\log 6}{\log 10} \times \frac{\log 3}{\log 6}-\log _{10}^{6}-\frac{\log 6}{\log 10} \cdot \frac{\log 7}{\log 6}

=\log _{10}^{5}+\log _{10}^{3}+\frac{\log 10 \times 3}{\log 10}+\frac{1}{2} \times 4 \log _{10}^{2}+\frac{1}{2} \times 4 \log _{10}^{2} \cdot \frac{1}{2} \log _{2} \frac{\log 3}{\log 10}-\log _{10}^{6}-\frac{\log 7}{\log 10}

=\log _{10}^{ \frac{10}{2} }+\log _{10}^{3} \frac{\log 10}{\log 10}+\frac{\log 3}{\log 10}+2 \log _{10}^{2}+\frac{\log 2}{\log 10} \times \frac{\log 7}{\log 2} \frac{\log 3}{\log 10}-\log _{10}^{23}-\frac{\log 7}{\log 10}

=\log _{10}^{10}-\log _{10}^{2}+\log _{10}^{3}+1+\log _{10}^{3}+2 \log _{10}^{2}+\log _{10}^{7}-\log _{10}^{3}-\log _{10}^{2}-\log _{10}^{3}-\log _{10}^{7}

=1-2 \log _{10}^{2}+2 \log _{10}^{2}+1-2 \log _{10}^{3}+2 \log _{10}^{3}+\log _{10}^{7}-\log _{10}^{7}

= 1 + 1

= 2

\therefore \text { L.H.S = R.H.S( proved })

(iii) \log _{2} \log _{2} \log _{4}^{256}+2 \log _{\sqrt{2}}^{2}=5

Solution:

\text { L.H.S }=\log _{2} \log _{2} \log _{4}^{256}+2 \log _{\sqrt{2}}^{2}

= \log _{2} \log _{2} \log _{4}^{4^{+}}+2 \log _{2 \sqrt{2}}^{2}

= \log _{2} \log _{2} 4 \log _{4}^{4}+\frac{2}{\frac{1}{2}} \log _{2}^{2}

= \log _{2} \log _{2}^{4}+4 \log _{2}^{2}

= \log _{2} \log _{2}^{2^{2}}+4 \times 1

= \log _{2}^{2} \log _{2}^{2}+4 \\= 1+4[\therefore \log _{2}^{2}=1]

= 5

\therefore \text { L.H.S }=\text { R.H.S (proved })

(iv) \log _{\mathrm{x}^{2}}^{\mathrm{x}} \times \log _{\mathrm{y}^{2}}^{\mathrm{y}} \times \log _{\mathrm{z}^{2}}^{z}=\frac{1}{8}

Solution:

\text { L.H.S }=\log _{x^{2}}^{x} \times \log _{y^{2}}^{y} \times \log _{z^{2}}^{z}

=\frac{1}{2} \log _{x}^{x} \times \frac{1}{2} \log _{y}^{y} \times \frac{1}{2} \log _{z}^{z}

=\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}

=\frac{1}{8}L.H.S = R.H.S ( proved )

(v) \log _{b^{3}}^{\mathrm{a}} \times \log _{\mathrm{c}^{3}}^{\mathrm{b}} \times \log _{\mathrm{a}^{3}}^{\mathrm{c}}=\frac{1}{27}.

Solution:

L.H.S =\log _{b^{3}}^{a} \times \log _{c^{3}}^{b} \times \log _{a^{3}}^{c}

=\frac{1}{3} \log _{b}^{a} \times \frac{1}{3} \log _{c}^{b} \times \frac{1}{3} \log _{a}^{c}

=\frac{1}{3.3 .3} \cdot \frac{\log a}{\log b} \cdot \frac{\log b}{\log c} \cdot \frac{\log c}{\log a}

=\frac{1}{27}

\therefore L.H.S = R.H.S (proved)

(vi) \frac{1}{\log _{x y}^{(x y z)}}+\frac{1}{\log _{y z}^{(x y z)}}+\frac{1}{\log _{z x}^{(x y z)}}=2

Solution:

\text { L.H.S }=\frac{1}{\log _{x y z}^{(x y z)}}+\frac{1}{\log _{y z}^{(x y z)}}+\frac{1}{\log _{z x}^{(x y z)}}

=\frac{1}{\frac{\log x y z}{\log x y}}+\frac{1}{\frac{\log x y z}{\log y z}}+\frac{1}{\frac{\log x y z}{\log z x}}

=\frac{\log x y}{\log x y z}+\frac{\log y z}{\log x y z}+\frac{\log z x}{\log x y z}

=\frac{\log x y+\log y z+\log z x}{\log x y z}

=\frac{\log x+\log y+\log y+\log z+\log z+\log x}{\log x+\log y+\log z}

=\frac{2 \log x+2 \log y+2 \log z}{\log x+\log y+\log z}

=\frac{2(\log x+\log y+\log z)}{\log x+\log y+\log z}

= 2

\therefore \text { L.H.S }=\text { R.H.S (proved })

(vii) \log \frac{a^{2}}{b c}+\log \frac{b^{2}}{c a}+\log \frac{c^{2}}{a b}=0

Solution:

\text { LH.S }=\log \frac{a^{2}}{b c}+\log \frac{b^{2}}{c a}+\log \frac{c^{2}}{a b}

= \log a^{2}-\log b c+\log b^{2}-\log c a+\log c^{2}-\log a b

= 2 \log a-\log b-\log c+2 \log b-\log c-\log a+2 \log c-\log a-\log b

= 2 \log a-2 \log a+2 \log b-2 \log b+2 \log c-2 \log c

= 0

\therefore \text { LH.S }=\text { RH.S(proved) }

(viii) x^{\log\text{y} +\log\text{z}} \times y^{\log\text{z} - \log \text{x}} \times z^{\log\text{x} +\log\text{y}}=1

Solution:

\text { Let } \mathrm{p}=x^{\log y-\log z} \times y^{\log z-\log x} \times z^{\log x-\log y}

Taking log both sides

\log p=\log \{x^{\log y-\log z} \times y^{\log z-\log x} \times z^{\log x-\log y}\}

\log p=\log x^{\log y-\log z}+\log y^{\log z-\log x}+\log z^{\log x-\log y}

\log p=(\log y-\log z) \log x+(\log z-\log x) \log y+(\log x-\log y) \log z

\log p=\log x \cdot \log y-\log x \cdot \log z+\log y \cdot \log z-\log y \cdot \log x+\log z \cdot \log x-\log z \cdot \log y

\therefore \log p=0

\text { or } \log p=\log 1[\therefore \log 1=0]

\therefore x^{\log y-\log z} \times y^{\log z-\log x} \times z^{\log x-\log y}=1(\text { proved })

6. (i) If \log \frac{x+y}{5}=\frac{1}{2}(\log x+\log y), \text{then let us show that} \frac{x}{y}+\frac{y}{x}=23

(ii) If \mathrm{a}^{4}+\mathrm{b}^{4}=14 \mathrm{a}^{2} \mathrm{~b}^{2}, \text{then let us show that} \log (\mathrm{a}^{2}+\mathrm{b}^{2})=\log a+\log b+2 \log 2

Solution:

(i) \therefore \log \frac{x+y}{5}=\frac{1}{2}(\log x+\log y)

\text { or, } \log \frac{x+y}{5}=\frac{1}{2}(\log x y)

\text { or, } \log \frac{x+y}{5}=\log (x y)^{ \frac{1}{2} }

\text { or, } \frac{x+y}{5}=(x y)^{\frac{1}{2}}

Squaring both sides,

(\frac{x+y}{5})^{2}=\{(x y)^{\frac{1}{2}}\}^{2}

\text { or, } \frac{x^{2}+2 x y+y^{2}}{25}=x y

\text { or, } x^{2}+2 x y+y^{2}=25 x y

\text { or, } x^{2}+y^{2}=25 x y-2 x y

\text { or, } x^{2}+y^{2}=23 x y

\text { or, } \frac{x^{2}+y^{2}}{x y}=23

\therefore \frac{x}{y}+\frac{y}{x}=23 \text { (proved) }

(ii) Given, a^{4}+b^{4}=14 a^{2} b^{2}

\text { or, } a^{4}+2 a^{2} b^{2}+b^{4}=14 a^{2} b^{2}+2 a^{2} b^{2}

\text { or, }(a^{2})^{2}+2 \cdot a^{2} \cdot b^{2}+(b^{2})^{2}=16 a^{2} b^{2}

\text { or, }(a^{2}+b^{2})^{2}=(4 a b)^{2}

\text { or, }(a^{2}+b^{2})=4 ab

Taking logarithm both side.

\text { or, } \log (a^{2}+b^{2})=\log (4 a b)

\text { or, } \log (a^{2}+b^{2})=\log 4+\log a+\log b

\text { or, } \log (a^{2}+b^{2})=\log 2^{2}+\log a+\log b

\therefore \log (a^{2}+b^{2})=\log a+\log b+2 \log 2 \text { (proved) }

7. If \frac{\log x}{y-z}=\frac{\log y}{z-x}=\frac{\log z}{x-y}, then let us show that xyz = 1

Solution:

\qquad \frac{\log x}{y-z}=\frac{\log y}{z-x}=\frac{\log z}{x-y}=k

\text { or, } \log x=k(y-z)=k y-k z

\text { or, } \log y=k(z-x)=k z-k x

\text { or, } \log z=k(x-y)=k x-k y

\therefore \log x+\log y+\log z=k y-k z+k z-k x+k x-k y

\text { or, } \log x y z=0

\text { or, } \log x y z=\log 1[\therefore \log 1=0]

\therefore x y z=1 \text { (proved) }

8. If \frac{\log x}{b-c}=\frac{\log y}{c-a}=\frac{\log z}{a-b}, then let us show that

(a) x^{b+c} \cdot y^{c+a} \cdot z^{a+b}=1

(b) x^{b^{2}+b c+c^{2}} \cdot y^{c^{2}+c a+a^{2}} \cdot z^{a^{2}+a b+b^{2}}=1

Solution:

(a) \frac{\log x}{b-c}=\frac{\log y}{c-a}=\frac{\log z}{a-b}=k(\text { say })

\log x=k(b-c)

\log y=k(c-a) \\\log z=k(a-b)

\therefore \text { Let } p=x^{b+c} \cdot y^{c+a} \cdot z^{a+b}

Taking logarithm both sides,

\log p=\log \cdot(x^{b+c} \cdot y^{c+a} \cdot z^{a+b})

=\log x^{b+c}+\log x^{c+a}+\log z^{a+b}

=(b+c) \log x+(c+a) \log x+(a+b) \log z

=(b+c) \cdot k(b-c)+(c+a) \cdot k(c-a)+(a+b) \cdot k \cdot(a-b)

=k(b^{2}-c^{2}+c^{2}-a^{2}+a^{2}-b^{2})=k .0=0

\therefore \log p=0

\text { or, } \log p=\log 1

\therefore p = 1

\therefore x^{b+c} \cdot y^{c+a} \cdot z^{a+b}=1 \text { ( proved })

(b) Let p=x^{b^{2}+b c+c^{2}} \cdot y^{c^{2}+c a+a^{2}} \cdot z^{c^{2}+a b+b^{2}}

Taking logarithm both sides

\log p=\log \{x^{b^{2}+b c+c^{2}} \cdot y^{c^{2}+c a+a^{2}} \cdot z^{a^{2}+a b+b^{2}}\}

\log p=\log x^{b^{2}+b c+c^{2}}+\log y^{c^{2}+c a+a^{2}}+\log z^{a^{2}+a b+b^{2}}

\log p=(b^{2}+b c+c^{2}) \log x+(c^{2}+c a+a^{2}) \log y+(a^{2}+a b+b^{2}) \log z

\log p=(b^{2}+b c+c^{2}) k \cdot(b-c)+(c^{2}+c a+a^{2}) k \cdot(c-a)+(a^{2}+a b+b^{2}) k(a-b)

\log p=k(b^{3}-c^{3}+c^{3}-a^{3}+a^{3}-b^{3})

\quad=k \times 0=0

\therefore \log p=0

\quad \log p=\log 1[\therefore \log 1=0]

\therefore p = 1

\therefore x^{b^{2}+b c+c^{2}} \cdot y^{c^{2}+c a+a^{2}} \cdot z^{c^{2}+a b+b^{2}}=1 \text { (Proved) }

9. If a^{3-x} \cdot b^{5 x}=a^{5+x} \cdot b^{3 x}, then let us show that, x \log (\frac{b}{a})=\log a

Solution:

Given, a^{3-x} \cdot b^{5 x}=a^{5+x} \cdot b^{3 x}

or, \frac{b^{5 x}}{b^{3 x}}=\frac{a^{5+x}}{a^{3-x}}

or, b^{5 x-3 x}=a^{5+x-(3-x)}

or, b^{2 x}=a^{5+x-3+x}

or, b^{2 x}=a^{2+2 x}

or, b^{2 x}=a^{2} \cdot a^{2 x}

or, \frac{b^{2 x}}{a^{2 x}}=a^{2}

or, (\frac{b}{a})^{2 x}=a^{2}

Taking logarithm both sides,

\text { or }, \log (\frac{b}{a})^{2 x}=\log a^{2}

\text { or }, 2 x \log (\frac{b}{a})=2 \log a

\therefore x \log (\frac{b}{a})=\log a(\text { proved })

10. let us solve:

(a) \log _{8}[\log _{2}\{\log _{3}(4^{x}+17)\}]=\frac{1}{3}

(b) \log _{8}^{x}+\log _{4}^{x}+\log _{2}^{x}=11

Solution:

(a) \log _{8}[\log _{2}\{\log _{3}(4^{x}+17)\}]=\frac{1}{3}

\text { or }, \log _{2}\{\log _{3} 4^{x}+17)\}=8^{\frac{1}{3}}=(2^{3})^{\frac{1}{3}}=2^{3 \times \frac{1}{3}}

\text { or, } \log _{2}\{\log _{3}(4^{x}+17)\}=2

\text { or }, \log _{3}(4^{x}+17)=2^{2}

\text { or, } \log _{3}(4^{x}+17)=4

\text { or, } 4^{x}+17=3^{4}=81

\text { or }, 4^{x}=81-17

\text { or, } 4^{x}=64

\text { or }, 4^{x}=4^{3}

\therefore x = 3 (Ans)

(b) \log _{8}^{x}+\log _{4}^{x}+\log _{2}^{x}=11

\text { or, } \log _{2^{x}}^{x}+\log _{2^{2}}^{x}+\log _{2}^{x}=11

\text { or, } \frac{1}{3} \log _{2}^{x}+\frac{1}{2} \log _{2}^{x}+\log _{2}^{x}=11

\text { or, }(\frac{1}{3}+\frac{1}{2}+1) \log _{2}^{x}=11

\text { or, }(\frac{2+3+6}{6}) \log _{2}^{x}=11

\text { or, } \frac{11}{6} \log _{2}^{x}=11

\text { or, } \log _{2}^{x}=6

\text { or, } x=2^{6}

\therefore x = 64 (Ans)

11. Let us show that the value of \log _{10}^{2} lies between \frac{1}{4} \ and \ \frac{1}{3}

Solution:

Let x=\log _{10}^{2}

10^{x}=2

L.C.M of denominator of 4 and 3 = 12

\therefore 10^{x}=2

(10^{x})^{12}=2^{12}=4096

\because 1000<4096<10000

or, 10^{3}<10^{12 x}<10^{4}

or, 3<12 x<4

or, \frac{3}{12}<x<\frac{4}{12}

or, \frac{1}{4}<x<\frac{1}{3}

or, \frac{1}{4}\log _{10}^{2}<\frac{1}{3}

Hence, the value of \log _{10}^{2} lies between \frac{1}{4} \ and \ \frac{1}{3}

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12. (M.C.Q)

(i) If \log _{\sqrt{x}}^{0.25}=4 then the value of x

(a) 0.5

(b) 0.25

(c) 4

(d) 16

Solution:

\log _{\sqrt{x}}^{0.25}=4

\text { or, } 0.25=(\sqrt{x})^{4}

\text { or, } \frac{25}{100}=x^{\frac{1}{2} \times 4}

\text { or, } \frac{1}{4}=x^{2}

\text { or, } x=\sqrt{\frac{1}{4}}=\frac{1}{2}

\therefore x = 0.5

(a) is correct answer

(ii)If \log _{10}^{(7 x-5)}=2, then the value of x

(a) 10

(b) 12

(c) 15

(d) 18

Solution:

\log _{10}^{7 x-5}=2

or, 7 x-5=10^{2}=100

or, 7 x=100+5

\text { or, } x=\frac{105}{7}=15

\therefore (c) is correct answer

(iii) If \log _{2}^{3}=a, then the value of \log _{8}^{27} is

(a) 3a

(b) \frac{1}{a}

(c) 2a

(d) a

Solution:

\log _{8}^{27}

= \log _{2^{3}}^{3^{3}}

= \frac{3}{3} \log _{2}^{3}

= \log _{2}^{3}

= a

\therefore(d) is correct answer

(iv) If \log _{\sqrt{2}}^{x}=a, then the value of \log _{2 \sqrt{2}}^{x} is

(a) \frac{a}{3}

(b) a

(c) 2a

(d) 3a

Solution:

\text { Now } \log _{\sqrt{2}}^{x}=a

=\log _{2^{\frac{1}{2}}}^{x}=a

=\frac{1}{2} \log _{2}^{x}=a

=\log _{2}^{x}=2 a

\because \log _{2 \sqrt{2}}^{x}

or, \log _{2^{ \frac{1}{2} }}^{x}

or, \frac{3}{2} \log _{2}^{x}

or, \frac{3}{2} \times 2 a=3 a

\therefore(d) is correct answer

(v) If \log _{x}^{\frac{1}{3}}=-\frac{1}{3}, then the value of x is

(a) 27

(b) 9

(c) 3

(d) \frac{1}{27}

Solution:

\log _{x}^{\frac{1}{3}}=-\frac{1}{3}

\text { or, } x^{-\frac{1}{3}}=\frac{1}{3}

\text { or, } \frac{1}{x^{\frac{1}{3}}}=\frac{1}{3}

\text { or, } x^{\frac{1}{3}}=3

\text { or, }(x^{\frac{1}{3}})^{3}=(3)^{3}

or, x = 27

\therefore (a) is acorrect answer

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13. Short answer type :

(i) Let us calculate the value of \log _{4} \log _{4} \log _{4}^{256}

Solution:

\log _{4} \log _{4} \log _{4}^{256}

= \log _{4} \log _{4}(\log _{4}^{4^{4}})

= \log _{4} \log _{4}(4 \log _{4}^{4})

= \log _{4} \log _{4}^{4}

= \log _{4}^{1}

= 0[\because \log _{4}^{1}=0]

(ii) Let us calculate the value of \log \frac{a^{n}}{b^{n}}+\log \frac{b^{n}}{c^{n}}+\log \frac{c^{n}}{a^{n}}

Solution:

\log \frac{a^{n}}{b^{n}}+\log \frac{b^{n}}{c^{n}}+\log \frac{c^{n}}{a^{n}}

= \log a^{n}-\log b^{n}+\log b^{n}-\log c^{n}+\log c^{n}-\log a^{n}

= 0 (Ans)

(iii) Let us show that a^{\log _{9}^{2}}=x

Solution:

Let p=\log _{a}^{x}

\text { or, } x =a^{p}

\text { or, } x =a^{\log _{a}^{x}}

\therefore a^{\log _{a}^{x}}

=x \quad \text { (proved) }

(iv) If \log _{e}^{2} \cdot \log _{x}^{25}=\log _{18}^{16} \cdot \log _{e}^{10} then let us calculate the value of x.

Solution:

\log _{e}^{2} \cdot \log _{x}^{25}=\log _{10}^{16} \cdot \log _{e}^{10}

\text { or, } \frac{\log 2}{\log e} \cdot \frac{\log 25}{\log x}=\frac{\log 16}{\log 10} \cdot \frac{\log 10}{\log e}

\text { or, } \log 2 \cdot \frac{\log 5^{2}}{\log x}=\log 2^{4}

\text { or, } \log 2 \cdot \frac{2 \log 5}{\log x}=4 \log 2

\text { or, } \frac{2 \log 5}{\log x}=4

\text { or, } 2 \log x=\log 5

\text { or, } \log x^{2}=\log 5

\text { or, } x^{2}=5

\therefore x=\sqrt{5}(\text { Ans })