Q.16^(a^2 + b )+ 16^(b^2 + a) = 1 find 36|a|+12|b|

asked by guest
on Sep 21, 2024 at 8:37 am



You asked:

Given \(Q \cdot {.16}^{{a}^{2} + b} + {16}^{{b}^{2} + a} = 1\) evaluate the expression \(36 \cdot \left\lvert a\right\rvert + 12 \cdot \left\lvert b\right\rvert\).

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