EXERCISE 1.6
EVIDYARTHIQ 1 Ex 1.6, Page No 26, Number Systems, Class 9th Maths
MKRLaws of Powers and Exponents
Class - 9th, Ex - 1.6, Q 1 ( NUMBER SYSTEM ) CBSE NCERT
Page No 26:
Question 1:
Find:(i) $64^{\frac{1}{2}}$
(ii) $32^{\frac{1}{5}}$
(iii) $125^{\frac{1}{3}}$
Answer:
(i) $64^{\frac{1}{2}}=\left(2^{6}\right)^{\frac{1}{2}}$$=2^{6 \times \frac{1}{2}}$ $\left[\left(a^{m}\right)^{n}=a^{m n}\right]$
$=2^{3}=8$
(ii) $32^{\frac{1}{5}}=\left(2^{5}\right)^{\frac{1}{5}}$
$=(2)^{5 \times \frac{1}{5}}$ $\left[\left(a^{m}\right)^{n}=a^{m n}\right]$
$=2^{1}=2$
(iii) $(125)^{\frac{1}{3}}=\left(5^{3}\right)^{\frac{1}{3}}$
$=5^{3 \times \frac{1}{3}}$ $\left[\left(a^{m}\right)^{n}=a^{m n}\right]$
$=5^{1}=5$
EVIDYARTHI
Q 2, Ex 1.6, Page No 26, Number Systems, CBSE Maths Class 9th
MKR
Class - 9th, Ex - 1.6, Q 2 ( NUMBER SYSTEM ) CBSE NCERT
Question 2:
Q2. Find:(i) $9^{\frac{3}{2}}$
(ii) $32^{\frac{2}{5}}$
(iii) $16^{\frac{3}{4}}$
(iv) $125^{\frac{-1}{3}}$
Answer:
(i) $9^{\frac{3}{2}}=\left(3^{2}\right)^{\frac{3}{2}}$$=3^{2 \times \frac{3}{2}}$ $\left[\left(a^{m}\right)^{n}=a^{m n}\right]$
$=3^{3}=27$
(ii) $(32)^{\frac{2}{5}}=\left(2^{5}\right)^{\frac{2}{5}}$
$=2^{5 \times \frac{2}{5}}$ $\left[\left(a^{m}\right)^{n}=a^{m n}\right]$
$=2^{2}=4$
(iii) $(16)^{\frac{3}{4}}=\left(2^{4}\right)^{\frac{3}{4}}$
$=2^{4 \times \frac{3}{4}}$ $\left[\left(a^{m}\right)^{n}=a^{m n}\right]$
$=2^{3}=8$
(iv) $(125)^{\frac{-1}{3}}=\frac{1}{(125)^{\frac{1}{3}}}$ $\left[a^{-m}=\frac{1}{a^{m}}\right]$
$=\frac{1}{\left(5^{3}\right)^{\frac{1}{3}}}$
$=\frac{1}{5^{3 \times \frac{1}{3}}}$ $\left[\left(a^{m}\right)^{n}=a^{m n}\right]$
$=\frac{1}{5}$
EVIDYARTHI
Q 3, Ex 1.6, Page No 26, Number Systems, Class 9th Mathematics
MKR
Class - 9th, Ex - 1.6, Q 3 ( NUMBER SYSTEM ) CBSE NCERT
Question 3:
Simplify:(i) $2^{\frac{2}{3}} \cdot 2^{\frac{1}{5}}$
(ii) $\left(\frac{1}{3^{3}}\right)^{7}$
(iii) $\frac{11^{\frac{1}{2}}}{11^{\frac{1}{4}}}$
(iv) $7^{\frac{1}{2}} \cdot 8^{\frac{1}{2}}$
Answer:
(i) $2^{\frac{2}{3}} \cdot 2^{\frac{1}{5}}=2^{\frac{2}{3}+1}$ $\left[a^{m} \cdot a^{n}=a^{m+n}\right]$$=2^{\frac{10+3}{15}}=2^{\frac{13}{15}}$
(ii) $\left(\frac{1}{3^{3}}\right)^{7}=\frac{1}{3^{3 \times 7}}$ $\left[\left(a^{m}\right)^{n}=a^{m n}\right]$
$=\frac{1}{3^{21}}$
$=3^{-21}$ $\left[\frac{1}{a^{m}}=a^{-m}\right]$
(iii) $\frac{11^{\frac{1}{2}}}{11^{\frac{1}{4}}}=11^{\frac{1}{2}-\frac{1}{4}}$ $\left[\frac{a^{m}}{a^{n}}=a^{m-n}\right]$
$=11^{\frac{2-1}{4}}=11^{\frac{1}{4}}$
(iv) $7^{\frac{1}{2}} \cdot 8^{\frac{1}{2}}=(7 \times 8)^{\frac{1}{2}}$ $\left[a^{m} \cdot b^{m}=(a b)^{m}\right]$
$=(56)^{\frac{1}{2}}$
0 comments:
Post a Comment