Solve-for-real-numbers-16-x-4-x-1-x-4-x-2-x-1-x-8-x-1-65-

Question Number 164124 by HongKing last updated on 14/Jan/22

Solve for real numbers :

\frac{16^{x} + 4^{x} + 1^{x}}{4^{x} + 2^{x} + 1^{x}} = \frac{8^{x} + 1}{65}

Answered by MJS_new last updated on 14/Jan/22

\frac{t^{4} + t^{2} + 1}{t^{2} + t + 1} = \frac{t^{3} + 1}{65}

\frac{(t^{2} + t + 1) (t^{2} - t + 1)}{t^{2} + t + 1} = \frac{(t + 1) (t^{2} - t + 1)}{65}

1 = \frac{t + 1}{65}

t = 64

x = 8

Commented by HongKing last updated on 15/Jan/22

thank you so much my dear Sir