Solve for $X$ , if $X^{a^{3}} . X^{b^{3}} . X^{3 a b (a + b)} = (2^{5})^{25},$ and $a + b = 5$ .

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B 2
$X^{a^{3}} . X^{b^{3}} . X^{3 a b (a + b)} = (2^{5})^{25}$

$\Rightarrow$ $X^{a^{3}} . X^{b^{3}} . X^{3 a^{2} b + 3 a b^{2}} = (2)^{125}$

$\Rightarrow$ $X^{a^{3} + b^{3} + 3 a^{2} b + 3 a b^{2}} = (2)^{125}$

$\Rightarrow$ $X^{(a + b)^{3}} = (2)^{125}$

$\Rightarrow$ $X^{(5)^{3}} = (2)^{125}$

$\Rightarrow$ $X^{125} = (2)^{125}$

$∴$ $X = 2$

Suggest Corrections

Similar questions

x^{a^{3}} . x^{b^{3}} . x^{3 a (a + b)} = (2^{5})^{25}

a + b = 5

Δ (x) = ∣ ∣ ∣ ∣ \begin{matrix} a_{1} + x & b_{1} + x & c_{1} + x a_{2} + x & b_{2} + x & c_{2} + x a_{3} + x & b_{3} + x & c_{3} + x \end{matrix} ∣ ∣ ∣ ∣

\frac{1}{{(243)}^{x}} = {(729)}^{y} = 3^{3}

B = [\begin{matrix} 1 & 1 8 & 3 \end{matrix}]

X = B^{2} - 4 B

X [\begin{matrix} a b \end{matrix}] = [\begin{matrix} 550 \end{matrix}]

x

- 3 x + 5 < 101

Join BYJU'S Learning Program