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

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Solution

The correct option is A $2$
$N o w x^{a^{3}} . x^{b^{3}} . x^{3 a b (a + b)} = {(2^{5})}^{25} \to x^{a^{3} + b^{3} + 3 a^{2} b + 3 {a b}^{2}} = 2^{125} \to x^{{(a + b)}^{3}} = {(2^{5})}^{3} \to x^{5^{3}} = 2^{5^{3}}$
Comparing Power, we get $x = 2$

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Similar questions

X

X^{a^{3}} . X^{b^{3}} . X^{3 a b (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} ∣ ∣ ∣ ∣

2 x + 5 = 3 (x + 2)

x

Solve for x

$x^{2}$ - 3x - 10 = 0

Solve for x

$x^{2} - 3 x - 10 = 0$

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