| Book Name | : Ganit Prakash |
| Subject | : Mathematics (Maths) |
| Class | : 9 (Madhyamik/WB) |
| Publisher | : Prof. Nabanita Chatterjee |
| Chapter Name | : Internal And External Division Of Straight Line Segment (21th Chapter) |
Let us work out – 21.1
Let us see whether the value exists for the following:
(1) \log _{2}^{-7} \quad (2) \log _{5}^{0} \quad (3) \log _{-3}^{2} \quad (4) \log _{0}^{2} \quad (5) \log _{1}^{7}
Solution:
(1) does not exist
(2) does not exist
(3) does not exist
(4) does not exist
(5) exist
1. Let us evaluate
(i) \log _{2 \sqrt{3}}^{1728}
(ii) \log _{0.01}^{0.000001}
(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}
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
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 })
