log2(x^2-6x+8)/log2(x-8) < 1

asked by guest
on Nov 25, 2024 at 7:08 am



You asked:

Solve the inequality \( \frac{\log_{2}\left( {x}^{2} - 6 x + 8 \right)}{\log_{2}\left( x - 8 \right)} < 1 \) for the variable \(x\).

MathBot Answer:

\[8 \lt x \lt 9\]

+ Submit Question