Question

If $\frac{a^{2b-3}\times {\left(a^2\right)}^{b+1}}{{\left(a^4\right)}^{-3}}=(a^3{)}^3÷(a^6{)}^{-3}$, then find the value of $2b$.

• A. 8
• B. 24
• C. 4
• D. 16

Answer 1

The correct option is A 8

$\frac{a^{2b-3}\times {\left(a^2\right)}^{b+1}}{{\left(a^4\right)}^{-3}}=(a^3{)}^3÷(a^6{)}^{-3}$
$\Rightarrow \frac{a^{2b-3}\times {\left(a\right)}^{2\times (b+1)}}{{\left(a\right)}^{4\times -3}}=\frac{(a{)}^{3\times 3}}{(a{)}^{6\times -3}}$
$\Rightarrow \frac{a^{2b-3}\times a^{2b+2}}{a^{-12}}=\frac{a^9}{a^{-18}}$
$\Rightarrow a^{2b-3+2b+2-(-12)}=a^{9-(-18)}$
$\Rightarrow a^{4b-1+12}=a^{9+18}$
$\Rightarrow a^{4b+11}=a^{27}$

As bases are equal on both the sides, the powers can be equated.
$\Rightarrow 4b+11=27$
$\Rightarrow 4b=16$
​​​​​​​$\Rightarrow 2b=8$