Systems of logarithmic equations

In a system of two logarithmic equations with two unknowns the most useful method to find the solution is, often, reduction or elimination, both used in solving Systems of linear equations.

Example

$\begin{array}{rcl} \log x + \log y & = & 3 \\ \log x - \log y & = & 1 \end{array}}$ We can add both equations, to eliminate one of the variables ( $\log y$ ), so that we have an equation where $x$ is the only variable.

We know how to solve equations where the variable $x$ is inside the logarithm so we can solve for $x$ : $+ \begin{array}{r} \log x + \log y = 3 \\ \underset{―}{\log x - \log y = 1} \\ 2 \log x + 0 = 4 \end{array}} \Rightarrow 2 \log x = 4 \Rightarrow \log x^{2} = 4 \Rightarrow x^{2} = 10^{4} \Rightarrow x = 10^{2} = 100$ Once we know the value of $x$ we can substitute in the first equation in order to obtain an equation with one unkown, $\log y$ .

Again, we know how to solve these equations, so we can solve for $y$ and we obtain: $\log x + \log y = 3 \Rightarrow \log 100 + \log y = 3 \Rightarrow 2 + \log y = 3 \Rightarrow \log y = 1 \Rightarrow y = 10^{1} = 10$ So, the solution to the system is: $x = 100$ and $y = 10$ .

In the systems of logarithmic equations we also need to verify that the solutions are possible, although in this case we can see immediately that they are valid solutions.

There is another type of system of logarithmic equations in which only one of the equations is logarithmic, while the other is a normal equation with the same two unknowns.

In these situations the best thing to do is to try to get rid of the logarithms and apply any of the methods that we like to solve the system.

Example

The following system consists of a logarithmic equation and a linear one: $\begin{array}{rcl} \log x - \log y & = & 1 \\ x + 2 y & = & 24 \end{array}}$ The first thing that it is necessary to do is to get rid of the logarithms. To do so, we can apply the property by which the difference of logarithm is the logarithm of their division so we can obtain: $\log x - \log y = 1 \Rightarrow \log \frac{x}{y} = 1 \Rightarrow \frac{x}{y} = 10$

We then have the following equation that we can solve quite easily: $\begin{array}{rcl} \frac{x}{y} = 10 \\ x + 2 y = 24 \end{array}}$

It is a system of two linear equations with two unknowns. We can solve it by the replacement method, since it is easy to clear $x$ from the first equation: $x = 10 y$

Now the above mentioned expression is replaced in the second equation and it is solved: $x + 2 y = 24 \Rightarrow 10 y + 2 y = 24 \Rightarrow 12 y = 24 \Rightarrow y = \frac{24}{12} = 2$ Once we know the value of $y$ we can put it back to the first equation to obtain the value of $x$ : $x = 10 y \Rightarrow x = 10 \cdot 2 = 20$

The solution to the system is, therefore, $x = 20$ and $y = 2$ , and it is a valid solution since both are positive numbers.

Example

$\begin{array}{rcl} \log x + \log y & = & 2 \\ x - y & = & 21 \end{array}}$ As in the previous case, the simplest thing is to get rid of the logarithms and to operate with linear equations. It is necessary to apply the property of the product of logarithms in the first equation: $\log x + \log y = 2 \Rightarrow \log (x \cdot y) = 2 \Rightarrow x y = 10^{2} \Rightarrow x y = 100$

The obtained equation is equivalent to the logarithmic one, so we have an equivalent linear system: $\begin{array}{rcl} x y = 100 \\ x - y = 21 \end{array}}$ We can apply the replacement method to express $x$ in terms of $y$ using the second equation: $x = 21 + y$ And we now replace into the first equation, obtaining an equation of second degree in $y$ : $x y = 100 \Rightarrow (21 + y) y = 100 \Rightarrow 21 y + y^{2} = 100 \Rightarrow y^{2} + 21 y - 100 = 0$

We can now solve this equation by applying the formula: $y = \frac{- 21 \pm \sqrt{21^{2} - 4 \cdot 1 \cdot (- 100)}}{2 \cdot 1} = \frac{- 21 \pm \sqrt{441 + 400}}{2} = \frac{- 21 \pm \sqrt{841}}{2} = \frac{- 21 \pm 29}{2}$

So that the possible values for $y$ are: $\begin{array}{rcl} y & = & \frac{- 21 + 29}{2} = 82 = 4 \\ y & = & \frac{- 21 - 29}{2} = \frac{- 50}{2} = - 25 \end{array}$

We still need to check whether those are valid solutions or not since the logarithm cannot take negative values. Now, if $y = - 25$ we know that the first equation should be: $\log x + \log (- 25) = 2$ And the $\log (- 25)$ does not exist.

On the other hand, if we take $y = 4$ , we obtain a valid solution. We can use this value to obtain the value of $x$ , by substituting it in the second equation of the system: $x = 21 + y \Rightarrow x = 21 + 4 = 25$

We still need to check that this is a valid value for $x$ . But since it is a positive value there are no problems and we see that the solution to the system is $x = 25$ and $y = 4$ .