If x=\sqrt{4\sqrt{4\sqrt{4...}}}, then what is the value of x?

Explanation

Let x = \sqrt{4\sqrt{4\sqrt{4...}}}. By squaring both sides, we get x^2 = 4x. Since x \gt 0, dividing by x gives x = 4.

Related questions on Algebra

• If p + q + r = 0, then what is z^{\frac{p^2}{qr}} \times z^{\frac{q^2}{rp}} \times z^{\frac{r^2}{pq}} equal to ?
• What is the value of k for which (k^2 - 5k + 4)x^2 + (k^2 - 3k - 4)x + (k^2 - 4k) = 0 is an identity ?
• If \frac{a^2}{b^2 + c^2} = \frac{b^2}{c^2 + a^2} = \frac{c^2}{a^2 + b^2}, then what is the value of a^4 + b^4 + c^4 equal to ?
• If \frac{1}{x} = \frac{1}{p} + \frac{1}{q}, then what is \frac{pq}{p^2 - q^2}\left(\frac{x + p}{x - p} - \frac{x + q}{x - q}\right) equa...
• If (x - 5) is the HCF of x^2 - x - p and x^2 - qx - 10, then what is the value of (p + q) ?