Chapter 6.6, Problem 11E

To determine

To calculate: The equation for the value of $x$ .

Answer to Problem 11E

The required value of $x$ is $-\frac{5}{3}$ .

Explanation of Solution

Given Information:

The given equation is ${49}^{5x+2}={\left(\frac{1}{7}\right)}^{11-x}$ .

Formula Used Power rule of logarithmic functions;

  ${\log }_a{x}^p=p{\log }_{a}x$

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

Calculation:

Consider the equation is ${49}^{5x+2}={\left(\frac{1}{7}\right)}^{11-x}$

Simplify the above equation.

  $\begin{array}{c}{49}^{5x+2}={\left(\frac{1}{7}\right)}^{11-x}\\ {\left\{{\left(7\right)}^2\right\}}^{\left(5x+2\right)}={\left\{{\left(7\right)}^{-1}\right\}}^{\left(11-x\right)}\\ 7^{\left(10x+4\right)}=7^{\left(x-11\right)}\end{array}$

Take logarithm each side with base $10$ of the above equation $7^{\left(10x+4\right)}=7^{\left(x-11\right)}$ .

  ${\log }_{10}{7}^{\left(10x+4\right)}={\log }_{10}{7}^{\left(x-11\right)}$

Now, use power rule of logarithm ${\log }_a{x}^p=p{\log }_{a}x$ in the above equation.

  $\left(10x+4\right){\log }_{10}7=\left(x-11\right){\log }_{10}7$

Divide each side by ${\log }_{10}7$ in the above equation.

  $\begin{array}{c}\frac{\left(10x+4\right){\log }_{10}7}{{\log }_{10}7}=\frac{\left(x-11\right){\log }_{10}7}{{\log }_{10}7}\\ 10x+4=x-11\\ 10x-x=-11-4\\ 9x=-15\end{array}$

Simplify the above equation.

  $\begin{array}{c}9x=-15\\ x=-\frac{15}{9}\\ x=-\frac{5}{3}\end{array}$

Hence, the required value of $x$ is $-\frac{5}{3}$ .