Inequalities with logarithmic functions

Log 0 is not defined also log is for positive numbers. So we have to keep these two things in mind to solve these questions.

1. log643xx+1>13${\log }_{64}{\displaystyle \frac{3x}{x+1}>{\displaystyle \frac{1}{3}}}$
1) -4>x
(2) -4>x and x>-1
(3) -4<x<-1
(4)1<x<4
(5) x≤−4$x\le -4$ and x≥4$\mathrm{x}\ge 4$
Ans:
log643xx+1>13${\log }_{64}{\displaystyle \frac{3x}{x+1}>{\displaystyle \frac{1}{3}}}$
3xx+1<6413${\displaystyle \frac{3x}{x+1}<{64}^{{\displaystyle \frac{1}{3}}}}$
3xx+1<4${\displaystyle \frac{3x}{x+1}<4}$
3xx+1−4<0${\displaystyle \frac{3x}{x+1}-4<0}$
−x−4x+1<0${\displaystyle \frac{-x-4}{x+1}<0}$
x+4x+1>0${\displaystyle \frac{x+4}{x+1}>0}$

x∈(−∞,−4)∪(−1,∞)$x\in (-\mathrm{\infty },-4)\cup (-1,\mathrm{\infty })$
Hence option 2.

2. log0.4(x2−13.5)≥−1${\log }_{0.4}(x^2-13.5)\ge -1$
(1)−4≤x≤4$-4\le x\le 4$ 
(2) x≤5$x\le 5$
(3) x≥0$\mathrm{x}\ge 0$ 
(4) x≤−5andx≥2$\mathrm{x}\le -5\mathrm{a}\mathrm{n}\mathrm{d}\mathrm{x}\ge 2$
(5) x≤−4andx≥4$\mathrm{x}\le -4\mathrm{a}\mathrm{n}\mathrm{d}\mathrm{x}\ge 4$
Ans:
log0.4(x2−13.5)≥−1${\log }_{0.4}(x^2-13.5)\ge -1$
x2−13.5≥0.4(−1)$x^2-13.5\ge {0.4}^{(-1)}$
x2−13.5≥2.5$x^2-13.5\ge 2.5$
x2−16≥0$x^2-16\ge 0$
(x+4)(x−4)≥0$(x+4)(x-4)\ge 0$
x≤−4$x\le -4$ and x≥4$\mathrm{x}\ge 4$
Hence option 5.

3. log(x2−36)≤log(4x−11)$\log (x^2-36)\le \log (4x-11)$
(1)4≤x≤6.11916$4\le x\le 6.11916$ 
(2) x<-6
(3) 6<x<7.38
(4) x>6
(5) None of these
Ans:
log(x2−36)≤log(4x−11)$\log (x^2-36)\le \log (4x-11)$
log(x2−36)−log(4x−11)≤0$\log (x^2-36)-\log (4x-11)\le 0$
log(x2−364x−11)≤0$\log \left({\displaystyle \frac{x^2-36}{4x-11}}\right)\le 0$
x2−364x−11≤1${\displaystyle \frac{x^2-36}{4x-11}\le 1}$
x2−364x−11−1≤0${\displaystyle \frac{x^2-36}{4x-11}-1\le 0}$
x2−36−4x+114x−11≤0${\displaystyle \frac{x^2-36-4x+11}{4x-11}\le 0}$
x2−4x−254x−11≤0${\displaystyle \frac{x^2-4x-25}{4x-11}\le 0}$
(x−(2−29−−√))(x−(2+29−−√))4x−11≤0${\displaystyle \frac{(x-(2-\sqrt{29}))(x-(2+\sqrt{29}))}{4x-11}\le 0}$
x∈(−∞,2−29−−√)∪(114,2+29−−√)$x\in (-\mathrm{\infty },2-\sqrt{29})\cup \left({\displaystyle \frac{11}{4},2+\sqrt{29}}\right)$

We also know that x2−36≥0$x^2-36\ge 0$ and 4x−11>0$4\mathrm{x}-11>0$
x∈(−∞,6)∪(6,∞)$x\in (-\mathrm{\infty },6)\cup (6,\mathrm{\infty })$ and x>114$\mathrm{x}>{\displaystyle \frac{11}{4}}$
Combining all conditions we get,

x∈(6,2+29−−√)$x\in (6,2+\sqrt{29})$
Hence option 3.

4. 9x−10.3x+9≤0$9^x-{10.3}^x+9\le 0$
(1) 1<x<2
(2) 1<x<9
(3) 0≤x≤2$0\le x\le 2$
(4) 2≤x≤9$2\le x\le 9$
(5) None of these
Ans:
9x−10.3x+9≤0$9^x-{10.3}^x+9\le 0$
Let 3x=a$3^x=a$
a2−10a+9≤0$a^2-10a+9\le 0$
a∈[1,9]$a\in [1,9]$
x∈[0,2]$x\in [0,2]$
Hence option 3.