Chapter 9.5, Problem 64STP

To determine

To solve the equation and also check the solution

Answer to Problem 64STP

The solution of the equation is $a=4$

Explanation of Solution

Given information:

The given equation is,

  ${\log }_{10}a+{\log }_{10}\left(a+21\right)=2$

Formula used:

  ${\log }_{a}n+{\log }_{m}m={\log }_a\left(n\cdot m\right)$

Calculation:

The equation can be solved as:

  $\begin{array}{l}{\log }_{10}a+{\log }_{10}\left(a+21\right)=2\\ \Rightarrow {\log }_{10}\left[a\left(a+21\right)\right]=2\\ \Rightarrow a^2+21a={10}^2\left[\text{Definition of logarithm}\right]\\ \Rightarrow a^2+21a=100\\ \Rightarrow a^2+21a-100=0\\ \Rightarrow a^2+\left(25-4\right)a-100=0\\ \Rightarrow a^2+25a-4a-100=0\\ \Rightarrow a\left(a+25\right)-4\left(a+25\right)=0\\ \Rightarrow \left(a+25\right)\left(a-4\right)=0\left[\text{Factor}\right]\\ a+25=0\text{ or }a-4=0\left[\text{Zero product property}\right]\\ a=-25\text{ and }a=4\end{array}$

However, $a=-25$ is extraneous, since it renders the original equation undefined. The only actual solution is $a=4$

Now, check the solution

Left-hand-side of the equation is:

  $\begin{array}{l}{\log }_{10}a+{\log }_{10}\left(a+21\right)={\log }_{10}4+{\log }_{10}\left(4+21\right)\left[\text{Put the value of }a\right]\\ ={\text{log}}_{10}\left(4\cdot 25\right)\\ ={\text{log}}_{10}100=2\end{array}$

So, left-hand-side is equal to right-hand-side

Therefore, the solution is $a=4$