Log x + log (x+15) = 2

How is this solved?

Log x + log (x+15) = 2

How is this solved?

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North Providence, RI

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

10^{2} = X(X + 15)

Now we can solve for X

X^{2} + 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

Cedar Park, TX

The **composite** of a **function** and **its inverse**
is just the *original argume*nt (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 log*_{10}x) therefore it's
**inverse** is **10**^{x}.

In summary:

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)]}} = 10^{2 }*(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

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.

Mark D.

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John P.

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