Question

How do you solve $\log x + \log (x + 21) = 2$?

Answer 1

Hint: We will first use the fact that log (ab) = log a + log b, then we will use the fact that if ${\log _b}a = c$, then $a = {b^c}$. Thus, we will form a quadratic in x which can be solved using quadratics formula.

Complete step by step solution:
We are given that we are required to solve $\log x + \log (x + 21) = 2$. ……………(1)
Since, we know that we have a formula which states that:
$ \Rightarrow $log a + log b = log a.b
Replacing a by x and b by (x + 21), we will then obtain the following equation:-
$ \Rightarrow \log x + \log (x + 21) = \log \{ x(x + 21)\} $
Simplifying the right hand side of the above mentioned equation, we will then obtain the following equation:-
$ \Rightarrow \log x + \log (x + 21) = \log \left( {{x^2} + 21x} \right)$
Putting this in equation number 1, we will then obtain the following equation:-
$ \Rightarrow \log \left( {{x^2} + 21x} \right) = 2$
We can also write this as follows:-
$ \Rightarrow {\log _{10}}\left( {{x^2} + 21x} \right) = 2$
Now, since we know that we have a formula given by the following expression:-
If ${\log _b}a = c$, then $a = {b^c}$.
Replacing b by 10, a by ${x^2} + 21x$ and c by 2 in the above mentioned formula, we will then obtain the following expression with us:-
$ \Rightarrow {x^2} + 21x = {10^2}$
Simplifying this, we will then obtain the following expression:-
$ \Rightarrow {x^2} + 21x - 100 = 0$
Now, we know that the general quadratic equation is given by $a{x^2} + bx + c = 0$ and its roots are given by the following expression:-
$ \Rightarrow x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Comparing it, we have: a = 1, b = 21 and c = - 100
So, the roots are:
$ \Rightarrow x = \dfrac{{ - 21 \pm \sqrt {{{21}^2} - 4 \times 1 \times ( - 100)} }}{2}$
Simplifying the expressions, we will then obtain the following expression:-
$ \Rightarrow x = \dfrac{{ - 21 \pm \sqrt {841} }}{2}$
Simplifying the above equation further, we will then obtain the following expression:-
$ \Rightarrow x = \dfrac{{ - 21 \pm 29}}{2}$
Thus, the possible values of x are 4 and – 25.

But since logarithmic of negative functions is not defined, therefore, x = 4 is the answer.

Note:-
The students must notice that there is an alternate way to solve the quadratic equation given by the following expression:-
$ \Rightarrow {x^2} + 21x - 100 = 0$
Now, we can write this equation as:-
$ \Rightarrow {x^2} - 4x + 25x - 100 = 0$
Taking x common from first two terms and 25 common from last two terms, we will then obtain the following equation:-
$ \Rightarrow x\left( {x - 4} \right) + 25\left( {x - 4} \right) = 0$
Taking (x – 4) common, we will get:-
$ \Rightarrow \left( {x - 4} \right)\left( {x + 25} \right) = 0$
Thus, the possible values of x are 4 and – 25.
But since logarithmic of negative functions is not defined, therefore, x = 4 is the answer.