Question

find the domain and range of x​

Answer 1

We're given to solve,

$\longrightarrow\sqrt[3]{2^{\frac{3x-1}{x-1}}}<8^{\frac{x-3}{3x-7}}$

or,

$\longrightarrow\left(2^{\frac{3x-1}{x-1}}\right)^{\frac{1}{3}}<\left(2^3\right)^{\frac{x-3}{3x-7}}$

$\longrightarrow2^{\frac{3x-1}{x-1}\cdot\frac{1}{3}}<2^{3\cdot\frac{x-3}{3x-7}}$

$\longrightarrow2^{\frac{3x-1}{3(x-1)}}<2^{\frac{3(x-3)}{3x-7}}$

Taking log to the base 2,

$\longrightarrow\dfrac{3x-1}{3(x-1)}<\dfrac{3(x-3)}{3x-7}$

Muliply both sides by 3,

$\longrightarrow\dfrac{3x-1}{x-1}<\dfrac{9(x-3)}{3x-7}$

Taking RHS to LHS,

$\longrightarrow\dfrac{3x-1}{x-1}-\dfrac{9(x-3)}{3x-7}<0$

$\longrightarrow\dfrac{(3x-1)(3x-7)-9(x-3)(x-1)}{(x-1)(3x-7)}<0$

$\longrightarrow\dfrac{9x^2-24x+7-9(x^2-4x+3)}{(x-1)(3x-7)}<0$

$\longrightarrow\dfrac{9x^2-24x+7-9x^2+36x-27}{(x-1)(3x-7)}<0$

$\longrightarrow\dfrac{12x-20}{(x-1)(3x-7)}<0$

Divide by 4.

$\longrightarrow\dfrac{3x-5}{(x-1)(3x-7)}<0$

This inequality is solved by wavy curve method as follows.

$\setlength{\unitlength}{1mm}\begin{picture}(5,5)\thicklines\put(0,0){\line(1,0){60}}\multiput(15,0)(15,0){3}{\circle{1.5}}\put(15,-5){1}\put(27.5,-5){ \frac{5}{3} }\put(45,-5){ \frac{7}{3} }\multiput(0,-15)(45,15){2}{\qbezier(0,0)(7.5,7.5)(15,15)}\qbezier(15,0)(22.5,7.5)(30,0)\qbezier(30,0)(37.5,-7.5)(45,0)\end{picture}$

Hence the solution to our inequality is,

$\longrightarrow\underline{\underline{x\in\bigg(-\infty,\ 1\bigg)\cup\left(\dfrac{5}{3},\ \dfrac{7}{3}\right)}}$

Answer 2

$\huge\bf\fbox\red{Answer:-}$

We're given to solve,

$\longrightarrow\sqrt[3]{2^{\frac{3x-1}{x-1}}} < 8^{\frac{x-3}{3x-7}}$

or,

$\longrightarrow\left(2^{\frac{3x-1}{x-1}}\right)^{\frac{1}{3}} < \left(2^3\right)^{\frac{x-3}{3x-7}}$

$\longrightarrow2^{\frac{3x-1}{x-1}\cdot\frac{1}{3}} < 2^{3\cdot\frac{x-3}{3x-7}}$

$\longrightarrow2^{\frac{3x-1}{3(x-1)}} < 2^{\frac{3(x-3)}{3x-7}}$

Taking log to the base 2,

$\longrightarrow\dfrac{3x-1}{3(x-1)} < \dfrac{3(x-3)}{3x-7}$

Muliply both sides by 3,

$\longrightarrow\dfrac{3x-1}{x-1} < \dfrac{9(x-3)}{3x-7}$

Taking RHS to LHS,

$\longrightarrow\dfrac{3x-1}{x-1}-\dfrac{9(x-3)}{3x-7} < 0$

$\longrightarrow\dfrac{(3x-1)(3x-7)-9(x-3)(x-1)}{(x-1)(3x-7)} < 0$

$\longrightarrow\dfrac{9x^2-24x+7-9(x^2-4x+3)}{(x-1)(3x-7)} < 0$

$\longrightarrow\dfrac{9x^2-24x+7-9x^2+36x-27}{(x-1)(3x-7)} < 0$

$\longrightarrow\dfrac{12x-20}{(x-1)(3x-7)} < 0$

Divide by 4.

$\longrightarrow\dfrac{3x-5}{(x-1)(3x-7)} < 0$

This inequality is solved by wavy curve method as follows.

$\setlength{\unitlength}{1mm}\begin{picture}(5,5)\thicklines\put(0,0){\line(1,0){60}}\multiput(15,0)(15,0){3}{\circle{1.5}}\put(15,-5){1}\put(27.5,-5){ \frac{5}{3} }\put(45,-5){ \frac{7}{3} }\multiput(0,-15)(45,15){2}{\qbezier(0,0)(7.5,7.5)(15,15)}\qbezier(15,0)(22.5,7.5)(30,0)\qbezier(30,0)(37.5,-7.5)(45,0)\end{picture}$

Hence the solution to our inequality is,

$\longrightarrow\underline{\underline{x\in\bigg(-\infty,\ 1\bigg)\cup\left(\dfrac{5}{3},\ \dfrac{7}{3}\right)}}$