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

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.