If $(a^{3})^{4} = (a^{6})^{b},$ the value $b$ is .

Open in App

Solution

Given:
$(a^{3})^{4} = (a^{6})^{b}$

To Find:
Value of $b$

Now, $(a^{3})^{4} = (a^{6})^{b}$

Applying repeated exponentiation rule: $(a^{m})^{n} = a^{m \times n}$

$⟹ a^{3 \times 4} = a^{6 \times b}$
$⟹ a^{12} = a^{6 b}$
As the base $(a)$ for both L.H.S and R.H.S is same, the corresponding exponents must be the same for the above equation.

$∴ 12 = 6 b$
$⟹ 6 b = 12$
$⟹ \frac{6 b}{6} = \frac{12}{6}$
$⟹ b = \frac{12}{6} = \frac{6 \times 2}{6 \times 1} = 2$

Therefore, the value of $b$ is $2.$

Suggest Corrections

Similar questions

a^{7} = 3

\frac{{(a^{- 2})}^{- 3} \times {(a^{3})}^{4} \times {(a^{- 17})}^{- 1}}{a^{7}}

If a and b are distinct positive primes such that $\sqrt[3]{a^{6} b^{- 4}} = a^{x} b^{2 y}$ ,find x and y.

sin (A + B) = cos (A - B)

(A + B)

If (x + a) (x + b) = x^{2} + 5 x + 6, what is the value of (a - b) if (a - b) > 0 ?

Join BYJU'S Learning Program