Chapter 4, Problem 2P

Solution

To choose: For which equation is $4$ a solution of given choices?

An option(choice) $(C)$ is correct.

Given: $4$

Calculation:

Let us consider given the value $x=4$ .

(A)Now, substitute $x=4$ in the equation $e^{2\cdot x}+1=17$ ;

  $\begin{array}{c}{e}^{2\cdot 4}+1=17\\ e^8+1=17\\ 2980+1\ne 17\left[\begin{array}{c}e\approx 2.718\\ e^8\approx 2980\end{array}\right]\\ 2981\ne 17\end{array}$

Here, the equation does not satisfy at $x=4$ .

(B)Similarly substitute $x=4$ in the equation ${\log }_2\left(x+2\right)={\log }_{2}2x$ ;

  $\begin{array}{c}{\log }_2\left(x+2\right)={\log }_{2}2x\\ {\log }_2\left(4+2\right)\ne {\log }_{2}2\cdot 4\\ {\log }_{2}6\ne {\log }_{2}8\end{array}$

Here, the equation does not satisfy at $x=4$ .

(C) Again, substitute $x=4$ in the equation $3^{x-2}+1=10$ ;

  $\begin{array}{c}{3}^{4-2}+1=10\\ 3^2+1=10\\ 9+1=10\\ 10=10\end{array}$

Here, the equation satisfies at $x=4$

(D) Again similarly substitute $x=4$ in the equation $\ln \left(x+2\right)+\ln x=1$ ;

  $\begin{array}{c}\ln \left(x+2\right)+\ln x=1\\ \ln \left(4+2\right)+\ln 4\ne 1\\ \ln \left(6\right)+\ln \left(4\right)\ne 1\\ \ln \left(24\right)\ne 1\end{array}$

Thus,

  $3.178\ne 1$

Here, the equation does not satisfy at $x=4$ .

Conclusion:

Hence, these three options are wrong at the value of $x=4$ . Because only the equation $3^{x-2}+1=10$ is satisfied at that $x=4$ and an option $(C)$ is true.