Calculus logarithm question

Question

Log x + log (x+15) = 2

 

How is this solved?

Answer 1

I use two rules which you should memorize.

 

First, if A and B are numbers greater than zero, then log A + log B = log AB.

This rule combines two logs into one.

 

Second, if log X = Y, then by definition, 10^Y = X

This rule changes a log equation into an exponential equation.

 

So starting with log X + log (X + 15) = 2, we use the first rule to get

 

log [X(X + 15)] = 2

 

using the second rule, we get

 

10² = X(X + 15) 

 

Now we can solve for X

 

X² + 15X - 100 = 0

 

(X - 5) (X + 20) = 0

 

Therefore X = 5 or X = -20

 

The answer X = -20 is eliminated because if we use it in the original equation we get log (-20) which is undefined because the domain of the log function is X > 0.

 

So we have one answer X = 5

Answer 2

The composite of a function and its inverse is just the original argument (argument means input) and so your problem is solved by understanding that the inverse of a log function is the exponential function with the same base. The base of the common log is 10 (log X means log₁₀x) therefore it's inverse is 10^x.

 

In summary:

 

      10^log(x) = x

 

We therefore solve the problem as follows:

 

log x + log (x+15) = 2  

 

log [x(x+15)] = 2 (the sum of two common logs is the common log of the product of their arguments)

 

10^{{log [x(x+15)]}} = 10²  (since left and right side of equation are equal we plug each into the 10^x function)

 

x(x+15) = 100

 

x^2 + 15x = 100

 

x^2 + 15x - 100 = 0 (factor the quadratic)

 

(x+20) (x-5) = 0

 

        By zero product rule:

        x+20 = 0 or x-5 = 0

 

so

 

x = -20 or x = 5

 

since the log function doesn't exist for negative numbers (log -20 doesn't exist) the answer is x = 5.