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Solve for r 16=z^(2^(2r-3))

$16 = z^{2^{2 r - 3}}$

Step 1

Rewrite the equation as $z^{2^{2 r - 3}} = 16$ .

$z^{2^{2 r - 3}} = 16$

$[\begin{array}{c} x^{2} & \frac{1}{2} \\ \sqrt{π} & \int x d x \end{array}]$

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