Ncert solutions

# Question 1

Q1) Using laws of exponents, simplify and write the answer in exponential form:

(i) 3^2\times3^4\times3^8

(ii) 6^{15}\div6^{10}

(iii) a^3\times a^2

(iv) 7^x\times7^2

(v) \left(5^2\right)^3\div5^3

(vi) 2^5\times5^5

(vii) a^4\times b^4

(viii) \left(3^4\right)^3

(ix) \left(2^{20}\div2^{15}\right)\times2^3

(x)8^t\div8^2

Solution 1:

(i) \left[a^m\times a^n=a^{m+n}\right]

\left[3^{2+4+8}\right]

3^{14}

(ii) \left[a^m\div a^n=a^{m-n}\right]

=\left[6^{15-10}\right]

=6^5

(iii) \left[a^m\times a^n=a^{m+n}\right]

=\left[a^{3+2}\right]

=a^5

(iv) \left[a^m\times a^n=a^{m+n}\right]

=\left[7^{x+2}\right]

(v) \left[\left(a^m\right)^n=a^{mn}\right]

=5^6\div5^3

=5^3

(vi) a^m\times b^{m\ }=\left(a\times b\right)^m

\left[2\times5\right]^5

\left[10\right]^5

(vii) a^m\times b^{m\ }=\left(a\times b\right)^m

=\left[a\times b\right]^4

(viii) \left[\left(a^m\right)^n=a^{mn}\right]

=\left[3\right]^{12}

(ix) \left[2^5\right]\times2^3

=\left[2\right]^8

(x) \left[a^m\div a^n=a^{m-n}\right]

=\left[8\right]^{t-2}

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