﻿ How to solve logarithmic equations step by step. Solved exercises.

# How to solve logarithmic equations step by step. Solved exercises.

Now I’m going to explain how to solve logarithmic equations step by step.

In order to understand all the steps well, it is important that you master the properties of the logarithms perfectly. In addition, you must also know how to convert a number into a logarithm.

## How to solve logarithmic equations

Logarithmic equations are those in which the incognita appears within the logarithm, for example:

Since the incognita is inside the logarithm, it is not possible to clear it directly.

In order to solve this type of equations, we must leave only one logarithm in each member of the equation. In addition, each logarithm cannot be multiplied by any number.

Once we have only one logarithm on both sides of the equation, we can eliminate the logarithms and thus be able to clear the unknowns.

In the previous example, we already have a logarithm in each member, so we can eliminate the logarithms and we are left:

Which is a first-degree equation, in which we can clear the unknown without problems:

This example that we have just seen, is a very simple example, but unfortunately, few times you will find such simple logarithmic equations.

The most normal thing is that you find equations where you have several logarithms in each member, some multiplied by some number and also combined with terms without logarithms (numbers or incognites).

This is where the application of the properties of logarithms comes into play, which will help us to simplify the equation to get that we have a logarithm in each member and thus be able to eliminate them.

The best way to learn to solve logarithmic equations is to practice and practice. So let’s solve a few logarithmic equations step by step.

## Solved exercises of logarithmic equations

### Exercise 1:

We can’t eliminate logarithms because in the second member we have a 2 multiplying the logarithm.

Thanks to the following property:

We can pass the number that multiplies the logarithm as exponent and we are left:

Now we can eliminate the logarithms and clear the x:

### Exercise 2:

We have a sum of two logarithms in the first member and a number in the second member.

The sum of two logarithms can be simplified into one, thanks to the following logarithm property:

The first member would fit us:

On the other hand, thanks to property 3, I can convert any number into a logarithm:

To convert 2 to logarithm, I have to express it as a logarithm where the base and content are equal and the content is raised to 2. As I have a logarithm in base 10, the content of the logarithm will therefore be a 10 raised to 2:

After applying these two properties, the equation is left:

Now we can eliminate the logarithms:

And clear the x:

### Exercise 3:

In this equation, first of all, the 2 that is multiplying the first logarithm, we pass it as exponent. On the other hand, we convert 2 to logarithm, as in the previous example:

Now, in the first member, we convert the subtraction of logarithms into a single logarithm by applying the following property:

And we have left:

We already have a single logarithm in each member, so we can eliminate them:

We are going to operate in order to solve the equation we have left. We pass the denominator to the second member multiplying

We eliminate the parenthesis by multiplying:

And we pass all the terms to the first member:

We are left with a second-degree equation, the solutions of which are:

But be careful, you have to check in the original equation if the two solutions are valid. If either of these two solutions made a negative logarithm, the solution would not be valid.

In this case, both solutions are valid.

### Exercise 4:

In the first member we have a division of logarithms. There is no property that we can apply to simplify a logarithm division (not to be confused with property 5).

What we can do is pass the logarithm of the denominator to the second member by multiplying to 2.

And now we pass 2 as an exponent of the logarithm:

In this case, it is not convenient to convert the 2 in logarithm, because we would have a multiplication of logarithms and we do not have a property that we can apply to simplify it.

We eliminate logarithms and it remains:

We developed the remarkable product of the second member:

We pass all terms to the first member:

And we resolve, whose solutions are:

We check if both solutions are valid, substituting in the original equation. In this case the second solution is not valid, since it returns negative to the content of the logarithm of the denominator and the logarithms of a negative number do not exist:

Therefore, the solution is:

### Exercise 5:

First, we pass the numbers we have multiplying the logarithms as exponent and the 2 we convert it to logarithm:

Now I could follow several paths. I am going to pass the first logarithm of the second member subtracting the first member, to have two logarithms in each member and be able to apply the properties:

In the first member I apply the property of the sum of logarithms and in the second member the property of the subtraction:

I eliminate logarithms:

Simplified:

Y cleared the x:

Whose solutions are:

But -20 is not a solution because it makes the logarithm number negative, so the solution of the equation is:

### Exercise 6:

Logarithmic equations are not always going to be with logarithms in base 10. In this case, it is with logarithms in base 2, but the way to solve them is exactly the same.

We pass the 2 that is multiplying the first logarithm as exponent and convert the 3 to a logarithm. Note that now, as we have logarithms in base 2, the 3 will be a logarithm in base 2, from 2 elevated to 3 (same base of the logarithm and number of the logarithm and the 3 is the exponent):

We apply the subtraction property of logarithms in the first member:

And we eliminate logarithms:

We pass the x of the denominator by multiplying the second member:

We develop the remarkable product:

We pass all terms to the first member and resolve:

Whose solutions are:

In this case, the solution of 0.10 is not valid as it converts the first logarithm of the negative equation.

### System of logarithmic equations:

Finally, we are going to solve a system of two logarithmic equations with two unknowns.

Let’s clear the x of the first equation:

We apply the property of the sum of logarithms in the first member and the 3 of the second member we convert it to logarithm:

We eliminate logarithms:

And we clear the x:

Now let’s also clear x in the second equation:

We apply the property of the subtraction of logarithms in the first member and the 10 of the second member we convert it to logarithm:

We eliminate logarithms:

And we cleared the x:

We match both expressions in which we clear the x:

And in this expression we have left, we clear the y:

Whose solutions are:

The second solution is not valid, so we discard it.

The first solution of y, we substitute it in any of the expressions in which we clear the x, for example in the second expression:

Replace the y with a 10 and is left:

So the system solution is: