Chapter 6.6, Problem 52E

To determine

To calculate: the value of $x$ .

Answer to Problem 52E

The required value of $x$ is $x>-0.53$ .

Explanation of Solution

Given Information:

The given inequality is $e^{3x+4}>11$ .

Formula Used:

Power rule of logarithmic functions:

  ${\log }_a{x}^p=p{\log }_{a}x$

Where $a$ is the base, $x$ is the variable and $p$ is the power of variable.

Property of logarithm:

  ${\log }_{a}a=1$

Calculation:

Consider the inequality is $e^{3x+4}>11$ .

Take natural logarithm with base $e$ each side of the above inequality.

  ${\ln }_e{e}^{3x+4}>{\ln }_{e}11$

Use power rule of logarithm ${\log }_a{x}^p=p{\log }_{a}x$ in the above inequality.

  $\left(3x+4\right){\ln }_{e}e>{\ln }_{e}11$

Now, use property of logarithm ${\log }_{a}a=1$ in the above inequality.

  $\begin{array}{c}\left(3x+4\right)\left(1\right)>{\ln }_{e}11\\ 3x+4>{\ln }_{e}11\end{array}$

Simplify the above inequality.

Substitute $2.4$ for ${\ln }_{e}11$ in the above inequality.

  $\begin{array}{c}3x+4>2.4\\ 3x>2.4-4\\ 3x>-1.6\end{array}$

Divide each side by $3$ in the above inequality.

  $\begin{array}{c}\frac{3x}{3}>\frac{-1.6}{3}\\ x>-0.53\end{array}$

Hence, the required value of $x$ is $x>-0.53$ .