Question

Solve the inequalities:
log2(x+1) <= 3
-x 13 years ago

Answer 1

in real numbers?

Answer 2

Yes.

Answer 3

x+1>0 , x>-1
\[\log_2(x+1)\le3 \]
\[x+1 \le 2^3\]
\[x \le 7\]
so \[x \in (-1;7>\]
first inequality

Answer 4

First one checks with my answer.

Answer 5

second
\[x^2>-x\]and
\[x^2<2x+1\]
\[x^2+x>0 \]
x(x+1)>0 and x^2-2x-1<0
x^2-2x-1=(x+sqrt2-1)(x-sqrt2-1)


\[x \in ((1-\sqrt2);0)\cup(1;(1+\sqrt2))\]