Question

log_2 ⁡(x-1) ≤ log_4⁡ (3-x) ...

Answer 1

there may be a snap way to do this, but i think it requires the following work:
first rewrite using the base of 2 using the change of base formula to get
\[\log_2(x-1)<\frac{\log_2(3-x)}{\log_2(4)}=\frac{\log_2(3-x)}{2}\]
\[2\log(x-1)<\log(3-x)\]
\[log((x-1)^2)<\log(3-x)\] and since log base 2 is strictly increasing this means
\[(x-1)^2<3-x\]

Answer 2

then you still have to solve the quadratic inequality
\[x^2-2x+1\leq 3-x\]
\[x^2-x-2\leq 0\]
\[(x-2)(x+1)\leq 0\]
\[-1\leq x\leq 2\] and then you STILL have to check that the solutions are in the domain of your original inequality

Answer 3

domain of
\[\log_2(x-1)\] is
\[x>1\] and domain of
\[\log_4(3-x)\] is
\[x<3\] so your actual solution is the interval
\[(1,3)\] or
\[1<x<3\]

Answer 4

i got it... i just did not know how ti speculate this one.. no idea how to get to next sequence \[\log_{2} x-1 = 1/2 \log_{2} 3-x\].. thank you

Answer 5

oh wait i made a mistake!

Answer 6

solution is
\[1<x\leq2\] sorry