Chapter 6.6, Problem 67E

To determine

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

Answer to Problem 67E

The required value of $x$ is $2.725$ .

Explanation of Solution

Given Information:

The given equation is $2^{1+3}=5^{x-1}$ .

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.

Calculation:

Consider the equation is $2^{1+3}=5^{x-1}$ .

Rewrite the above equation.

  $2^4=5^{x-1}$

Take common logarithm each side with base $10$ .

  ${\log }_{10}{2}^4={\log }_{10}{5}^{x-1}$

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

  $4{\log }_{10}2=\left(x-1\right){\log }_{10}5$

Substitute $0.301$ for ${\log }_{10}2$ and $0.698$ for ${\log }_{10}5$ in the above equations.

  $4\left(0.301\right)=\left(x-1\right)\left(0.698\right)$

Simplify the above equation.

  $\begin{array}{l}4\left(0.301\right)=\left(x-1\right)\left(0.698\right)\\ 1.204=\left(0.698\right)x-0.698\\ 1.204+0.698=\left(0.698\right)x\\ 1.902=\left(0.698\right)x\end{array}$

Simplify the equation.

  $\begin{array}{c}1.902=\left(0.698\right)x\\ \frac{1.902}{0.698}=x\\ 2.7249=x\\ x\approx 2.725\end{array}$

Hence, the required value of $x$ is $2.725$ .