Logarithmic equations are fundamental in the study of mathematics, as they allow solving problems in various fields, from science to economics. A logarithm, in simple terms, answers the question: "To what exponent must we raise a base to obtain a given number?" This relationship converts multiplications into additions, facilitating the resolution of complex equations.

Solve the following logarithmic equations:

1

Solution

1 In the first member we apply the property of the logarithm of a quotient



2 Taking into account that



3 Taking into account the definition of logarithm and that it is a common logarithm:

2

Solution

1 In the first member we apply the property of the logarithm of a product

2 Taking into account that

3 Taking into account the definition of logarithm and that it is a common logarithm:

3

Solution

1 In the first member we apply the property of the logarithm of a quotient

2 Taking into account that

3 Taking into account the definition of logarithm and that it is a common logarithm:

4

Solution

1 In the first member we apply the property of the logarithm of a product

2 Taking into account that

3 Taking into account the definition of logarithm and that it is a common logarithm:

5

Solution

1 In the first member we write the root as a fractional power and apply the property of the logarithm of a power

2 We solve for the logarithm

3 Taking into account the definition of logarithm and that it is a common logarithm:

6

Solution

1 In the second member we apply the property of the logarithm of a quotient

2 We subtract from both members and taking into account that , we have:

3 Taking into account the definition of logarithm and that it is a common logarithm:

7

Solution

1 In the first member, we apply the property of the sum of logarithms:

2 Taking into account the injectivity of logarithms (or equating the arguments) we have:

3 We solve the equation and check the solution

8

Solution

1 We apply the property of the logarithm of a power in both members

2 We apply the property of the logarithm of a product

3 We perform operations in the first member

4 We apply the injectivity of logarithms to remove logarithms

5 We solve the equation

6 Neither nor are solutions because if we substitute them in the equation we encounter logarithm of 0 and logarithm of a negative number and such logarithms do not exist, so the only solution is

9

Solution

1 We move to the second member and apply the power property in both members

2 We apply the injective property and find the values of

3 Solving the first factor we obtain , which is an inconsistency and means the equation has no solution. Solving the second factor we have , but is not defined and means the equation has no solution.

10

Solution

1 We clear denominators and make a change of variable

2 Solving the equation

3 We undo the change of variable and apply the definition of logarithm

11

Solution

1 We move the second term to the 2nd member and apply the property of the logarithm of a power

2 We apply the injectivity of logarithms and develop the operations

3 We solve the equation applying the quadratic formula

12

Solution

1 We multiply both members by

2 In the second member we apply the property of the logarithm of a power and take into account the injectivity of logarithms

3 We solve the equation, is not a solution because we would encounter the logarithm of a negative number in the denominator when substituting in the equation.

13

Solution

1 We clear denominators

2 In the second member we apply the property of the logarithm of a power and subsequently apply the injectivity of logarithms

3 We perform the operations and solve the quadratic equation

14

Solution

1 In the first member we apply the logarithm of a product and in the second the property of the logarithm of a power.

2 Taking into account the injectivity of logarithms we have:

3 We solve the equation and verify that we do not obtain a zero or negative logarithm

15

Solution

1 We multiply both members by and move everything to the first member

2 Considering that and clearing denominators:

3 We make a change of variable

3 We solve the equation

4 We undo the change of variable

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Agostina Babbo

Agostina Babbo is an English and Italian to Spanish translator and writer, specializing in product localization, legal content for tech, and team sports—particularly handball and e-sports. With a degree in Public Translation from the University of Buenos Aires and a Master's in Translation and New Technologies from ISTRAD/Universidad de Madrid, she brings both linguistic expertise and technical insight to her work.