Chapter 8.6, Problem 7E

To determine

To find:the solution of inequality equation and verify the solution.

Answer to Problem 7E

  $\frac{2}{3}<x<\frac{1}{2}$ .

Explanation of Solution

Given:

  ${\log }_5\left(2x+1\right)>{\log }_4\left(3x-2\right).$

Concept used:

In exponential form using the definition of the logarithm. If $\text{x and b }$ are positive real numbers and $\text{b}$ does not equal to $\text{1}$ ,then:

  ${\text{log}}_{\text{b}}\left(\text{x}\right)\text{=y}$ which is equivalent to the ${\text{b}}^{\text{y}}\text{=x}$ .

  $\log x^a=a\log x.$

The value or domain inside the logarithm is always greater than $0$ .

Calculation:

According to the given:

  ${\log }_5\left(2x+1\right)>{\log }_4\left(3x-2\right).$

By putting $x=2$ .

  $\begin{array}{l}{\log }_5\left(2\left(2\right)+1\right)<{\log }_4\left(3\left(2\right)-2\right).\\ {\log }_5\left(5\right)<{\log }_4\left(4\right).\\ 1<1.\end{array}$

Since, the same number is greater than same number which is not possible but the value inside the logarithm must be greater than 0.

  $\begin{array}{l}2x+1>0\\ x>\frac{-1}{2}.\\ x<\frac{1}{2}.\\ 3x-2>0.\\ x>\frac{2}{3}.\end{array}$

Hence, the solution of the inequality equation is:

  $\frac{2}{3}<x<\frac{1}{2}$ .