I don't get how my teacher got two different equations out of the one. One is $> 0$ and the other one is $<2$. Be detailed please.
2026-04-07 12:30:25.1775565025
Solve $\log_2 (1+\frac{1}{x-1})<1$
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If your inequation is $$ \log_2\left(1+\frac{1}{x-1}\right)<1 $$ then it is equivalent to the system of inequations \begin{cases} 1+\dfrac{1}{x-1}<2\\[2ex] 1+\dfrac{1}{x-1}>0 \end{cases} because $2>1$ and so the exponential with base $2$ is increasing. The first is obtained by removing the logarithm, the second because you need to ensure the logarithm exists.