If $\frac{a^{2b-3}\times (a^2)^{b+1}}{( a^4)^{-3}}=(a^3)^3\p(a^6)^{-3}$. Find the value of $2b$.

Given: $\frac{a^{2b-3}\times (a^2)^{b+1}}{( a^4)^{-3}}=(a^3)^3\div(a^6)^{-3}$.

To do: To find the value of $2b$.

Solution:

$\frac{a^{2b-3}\times (a^2)^{b+1}}{( a^4)^{-3}}=(a^3)^3\div(a^6)^{-3}$

$\Rightarrow \frac{a^{2b-3}\times (a)^{2( b+1)}}{( a^4)^{-3}}=(a^{3\times3})\div(a^{6\times-3})$       [$\because ( x^m)^n=x^{mn}$]

$\Rightarrow \frac{a^{2b-3}\times (a)^{2( b+1)}}{( a^4)^{-3}}=(a^{3\times3})\div(a^{6\times-3})$

$\Rightarrow \frac{a^{2b-3}\times (a)^{( 2b+2)}}{( a)^{-12}}=(a^{9})\div(a^{-18})$

$\Rightarrow \frac{a^{( 2b-3+2b+2)}}{( a)^{-12}}=(a^{9})\times(a^{18})$    [$\because 1\div x^{-m}=1\times x^m$]

$\Rightarrow a^{( 4b-1)}\times( a)^{12}=a^{( 9+18)}$

$\Rightarrow a^{( 4b-1+12)}=a^{( 9+18)}$

$\Rightarrow a^{( 4b-11)}=a^{( 27)}$

$\Rightarrow 4b-11=27$     [$\because\ if\ a^m=a^n\Rightarrow m=n$]

$\Rightarrow 4b=27+11$

$\Rightarrow 4b=38$

$\Rightarrow b=\frac{38}{4}$

$\Rightarrow 2b=\frac{38}{4}\times2$

$\Rightarrow 2b=19$

Thus, the value of $2b$ is $19$.

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Updated on: 10-Oct-2022

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