Question

Find the value of 'a' satisfying the equation.
a) ${\left(4^2\right)}^a={\left(7^a\right)}^2$.
b) $(2^a{)}^5=2^4\times 4^3$
[4 MARKS]

Answer 1

Each option: 2 Marks

a) Given that,

${\left(4^2\right)}^a={\left(7^a\right)}^2$

$4^{2a}=7^{2a}$ $[{a^m}^n=a^{mn}]$

Given, $4^{2a}=7^{2a}$

Since $4\ne 7$ the only case where the above equation is true is when both the exponents are zero.

$\Rightarrow a=0$

$\Rightarrow 4^{2a}=7^{2a}=1$ $[\because 4^0=1,7^0=1]$


b) The given equation is

$(2^a{)}^5=2^4\times 4^3$

$\Rightarrow 2^{a\times 5}=2^4\times (2^2{)}^3$

$\because (a^m{)}^n=a^{m\times n}$

$\Rightarrow 2^{a\times 5}=2^4\times \left(2^{2\times 3}\right)$

$\Rightarrow 2^{a\times 5}=2^4\times \left(2^6\right)$

$\Rightarrow 2^{a\times 5}=2^{4+6}=2^{10}$

$\because a^m\times a^n=a^{m+n}$

Since their bases are same and they are equal, threfore their powers must be same,

So, $5\times a=10$

$\Rightarrow a=\frac{10}{5}$

$\Rightarrow a=2$

So, the value of a = 2.