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Calculus logarithm question

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2 Answers

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, 10Y = 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
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 10x.
 
In summary:
 
      10log(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 10x 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.