Logarithmic equations of first degree

In a logarithmic equation there are one or several unknowns affected by a logarithm.

Example

$\log x + 4 = 6$

To solve this type of equations it is necessary to bear in mind the properties of the logarithms, which are summarized next:

$\log_{a} x^{n} = n \cdot \log_{a} x$
$\log_{a} (x \cdot y) = \log_{a} x + \log_{a} y$
$\log_{a} (\frac{x}{y}) = \log_{a} x - \log_{a} y$

But also, it is necessary to bear in mind the following two rules related to the operations in equations:

$\log_{a} x = \log_{a} y \Rightarrow x = y$

Namely, if both members of an equation are under logarithms then they are the same

And, we must remember the logarithm definition: $\log_{a} x = b \Rightarrow x = a^{b}$

Or , the logarithm of base $a$ of $x$ equal $b$ , implies that $a$ raised to the power $b$ is the number $x$ .

Example

By just knowing these two rules it is possible to solve the equation $\log x + 4 = 6$ .

First of all, it is necessary to isolate $x$ on one side and the independent terms on the second one: $\log x = 6 - 4$ The second term is: $\log x = 2$ And the rule is applied: $x = 10^{2} = 100$

It is necessary to remember that, when we compute the power we have to keep the basis of the logarithm.

In the example, the equation contains a logarithm of base $10$ and, therefore, we have to keep the $10$ as the base for the power.

If the logarithm was in base $2$ , the power should keep the same base. If the base is not specified and the symbol used is $\ln$ or if we have the symbol $e$ , then it is the number $e$ that should be used as a base.

Example

$\log 4 x = \log 2 + 1$ First, we put the logarithms on one side and the constants on the other: $\log 4 x - \log 2 = 1$ By the property of the quotient of the logarithms the first term can be simplified in the following way: $\log \frac{4 x}{2} = 1$

Now the logarithm can be moved to the other side of the equality by doing: $\frac{4 x}{2} = 10^{1} \to \frac{4 x}{2} = 10$

At this point, the expression is a linear equation with a normal unknown and, therefore, simple to solve. We can simplify the $4$ and the $2$ and then obtain: $2 x = 10 \Rightarrow x = \frac{10}{2} = 5$

Example

$\log (2 x + 1) = \log x + 2$ An option to solve this equation is to follow the same steps used to solve the previous one: unknowns on the left side and independent terms on theright, so that: $\log (2 x + 1) - \log x = 2$

By the property of the quotient of the logarithms it is possible to group the first member so that: $\log (\frac{2 x + 1}{x}) = 2$

Now it is possible to send the logarithm to the other side of the equality, obtaining a linear equation with one unknown: $\frac{2 x + 1}{x} = 10^{2} \Rightarrow \frac{2 x + 1}{x} = 100 \Rightarrow 2 x + 1 = 100 x \Rightarrow 2 x - 100 x = - 1 \Rightarrow$ $\Rightarrow - 98 x = - 1 \Rightarrow x = \frac{1}{98}$

Another option to solve the same exercise is to try to express the constant term in terms of a logarithm of basis $10$ .

Then we can use the first rule we introduce to obtain the same linear equation with one unknown that we know how to solve. But: how to express $2$ as a logarithm? The best way to think about how to do it is by raising an equation and solving it: $\log x = 2$ Namely: what is the number that when we take the logarithm we obtain a $2$ ? For this, it is necessary only to clear $x$ , as we have been doing previously: $x = 10^{2} = 100$ Now, it is possible to raise an equation equivalent to the first one, but with all its members inside the logarithm ( $\log 100$ instead of $2$ ): $\log (2 x + 1) = \log x + \log 100$

We can use the rules for the sum of logarithms: $\log (2 x + 1) = \log (100 x)$ And, now the logarithms can be eliminated. We obtain a linear equation with one unkown, which is the same as we obtained previously: $2 x + 1 = 100 x \Rightarrow 2 x - 100 x = - 1 \Rightarrow - 98 x = - 1 \Rightarrow x = \frac{1}{98}$

It is important to note that when working with logarithms we have to take into account that they can only take positive numbers. Thus, some of the solutions obtained from the linear equation might not be valid.

Example

$\log (x - 7) - \log 2 x = 0$ In this case it is possible to try to eliminate the logarithms and to obtain an equivalent equation. We move one of the terms to the other side of the identity: $\log (x - 7) = \log (2 x)$

At this point, the logarithms can be eliminated, obtaining a linear sytem with one unknown: $x - 7 = 2 x \Rightarrow x - 2 x = 7 \Rightarrow - x = 7 \Rightarrow x = - 7$

So far everything seems correct, but when we put the result back in the initial logarithmic equation, we obtain the following: $\log (- 7 - 7) - \log (2 \cdot (- 7)) = 0 \Rightarrow \log (- 14) - \log (- 14) = 0$ The problem is that the logarithms need to take positive numbers to be well defined. So $x = - 7$ is not a possible solution.

In these cases we will say that the equation has no solution.