Question

How do you solve $\log x + \log (x + 15) = 2$ .

Answer 1

Hint : We need to find the value of ‘x’. To solve this we need to know the logarithm product rule, $\log (A) + \log (B) = \log (A.B)$ . After using this and applying antilog on both sides we will get a quadratic equation. We can solve the obtained equation by factorization method or using Sridhar Acharya’s formula that is $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ .

Complete step-by-step answer :
Given,
$\log x + \log (x + 15) = 2$ , here we make note that it is logarithmic base 10.
Using the logarithm product rule $\log (A) + \log (B) = \log (A.B)$ . Comparing we have, $A = x$ and $B = x + 15$ .
$\Rightarrow \log (x.(x + 15)) = 2$
Multiplying ‘x’ in the brackets we have,
$\Rightarrow \log ({x^2} + 15x) = 2$
Applying antilog on both sides we get,
$\Rightarrow {x^2} + 15x = {10^2}$
Rearranging we have,
$\Rightarrow {x^2} + 15x - 100 = 0$
Let’s use Sridhar Acharya’s formula. That is if we have an quadratic equation $a{x^2} + bx + c = 0$ then $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ .
Here, $a = 1$ , $b = 15$ and $c = - 100$ .
Substituting we have,
$x = \dfrac{{ - 15 \pm \sqrt {{{15}^2} - \left( {4 \times 1 \times ( - 100)} \right)} }}{{2 \times 1}}$
$= \dfrac{{ - 15 \pm \sqrt {225 - \left( {4 \times - 100} \right)} }}{2}$
$= \dfrac{{ - 15 \pm \sqrt {225 - \left( { - 400} \right)} }}{2}$
We know negative is multiplied by negative we get positive,
$= \dfrac{{ - 15 \pm \sqrt {225 + 400} }}{2}$
$= \dfrac{{ - 15 \pm \sqrt {625} }}{2}$
We know that square root of 625 is 25,
$= \dfrac{{ - 15 \pm 25}}{2}$
Thus we have two roots,
$= \dfrac{{ - 15 + 25}}{2}$ and $= \dfrac{{ - 15 - 25}}{2}$
$= \dfrac{{10}}{2}$ and $= \dfrac{{ - 40}}{2}$
$= 5$ and $= - 20$
Thus we have roots 5 and -20.
We know that the arguments of the logarithmic is positive.
So we ignore -20.
Hence the required solution is $x = 5$ .
So, the correct answer is “ $x = 5$ ”.

Note : We need to know the basic property of logarithms. That is the logarithm quotient rule: $\log \left( {\dfrac{A}{B}} \right) = \log (A) - \log (B)$ . Logarithm power rule: $\log ({A^B}) = B.\log (A)$ etc. in above we can solve the quadratic equation by factorization method. we have ${x^2} + 15x - 100 = 0$ , we can rewrite it as $\Rightarrow {x^2} + 20x - 5x - 100 = 0$
$\Rightarrow x(x + 20) - 5(x + 20) = 0$
Taking common $(x + 20)$ we have,
$\Rightarrow (x + 20)(x - 5) = 0$
Using the principle of zero products we have,
$\Rightarrow x + 20 = 0$ and $x - 5 = 0$
$\Rightarrow x = - 20$ and $x = 5$ . We can see that in both the cases we have the same values for ‘x’. Careful in the calculation part.