Logarithmic Equations

Exponents can also be written as logarithms. We know that 2⁵ equals 32. However, if we are asked to find what number replaces the question mark in 2? = 32, we can simply find the answer by trial and error. But what if the question mark is hidden in 2? = 30? Is there a number that when multiplied by two equals 30? If not, what is the solution? Logarithms are the answer (or logs).

Let’s look at some examples of logarithms, as well as their laws and attributes.

Definition:

The exponent by which b must be raised to get x is the logarithm of a positive real integer x with regard to base b. In other words, the unique real number y such that by=x is the logarithm of x to base b.

“logb x” is the notation for the logarithm (pronounced as “the logarithm of x to base b”, “the base-b logarithm of x”, or most commonly “the log, base b, of x”).

The function logb is the inverse function of the function x⇔bx, which is a more concise formulation.

Logarithmic equations:

A logarithmic equation is one in which the logarithm of an expression with a variable is used. To solve exponential equations, determine if both sides of the problem can be written as powers of the same integer.

Properties of logarithms:

Scientists immediately adopted logarithms due to a variety of advantageous qualities that made long, painstaking calculations easier. Scientists may, for example, find the product of two numbers m and n by looking up the logarithms of each number in a special table, adding the logarithms together, and then consulting the table again to find the number with the calculated logarithm (known as its antilogarithm). This relationship is expressed in terms of common logarithms as log mn = log m + log n. 100 1,000, for example, can be determined by looking up the logarithms of 100 and 1,000, putting the logarithms together , and then looking up the antilogarithm (100,000) in the table. Similarly, logarithms are used to convert division difficulties into subtraction problems: log m/n = log m log n. Not only that, but logarithms can be used to simplify the calculation of powers and roots. As illustrated in the example, logarithms can be converted between any positive bases (excluding 1 because all of its powers are equal to 1), as shown in the example. 

Products: log_b mn = log_bm + log_bn.

Ratios: log_b m/n = log_bm – log_bn.

Powers: log_b np = p log_bn.

Roots: log_b q√n = 1/q log_bn.

Change of bases: log_bn = logan log_ba.

How to Solve Logarithmic Equation:

The first step in solving a logarithmic equation like log2 (5x + 7) = 5 or log3 (7x + 3) = log3 (5x + 9) is deciding how to solve the problem. Some logarithmic problems can be addressed by eliminating the logarithms, while others must be rewritten in exponential form. How can we know which method is the best for solving a logarithmic problem? The trick is to examine the problem and determine whether it contains solely logarithms or terms that are not logarithms.

If we take the problem log2 (5x + 7) = 5 as an example. There is a term in this problem, 5, that does not have a logarithm. Rewriting the logarithmic problem in exponential form is the correct technique to tackle these types of logarithmic problems. This issue involves solely logarithms if we examine the example log₃ (7x + 3) = log₃ (5x + 9). So, to solve these types of logarithmic problems correctly, simply drop the Logarithms.

Logarithms examples:

Example1. Solve Log₃ (9x + 2) = 4.

Solution. Log₃ (9x + 2) = 4

9x + 2 = 3⁴

9x + 2= 81

x = 79/9

Therefore, the solution to the problem Log₃ (9x + 2) = 4 is x = 79/9.

Example2. Solve Log₄ x + log₄ (x -12) = 3

Solution. Log₄ x + log₄ (x -12) = 3

Log4 (x(x -12)) = 3

x(x-12)=4³

x² – 12x = 64

x²-12x-64=0

(x+4) (x-16) = 0

X=-4 or x= 16

X= 16

Therefore, the solution to the problem Log₄ x + log₄ (x -12) = 3 is x = 16.

Example3. Solve  log(5x -11) = 2

Solution. Log(5x -11) = 2

5x – 2 =10²

5x-2=100

x = 102/5

Therefore, the solution to the problem log(5x -11) = 2 is x= 102/5.

Example4. Solve  Log₂ (x +1) – log₂ (x – 4) = 3

Solution. Log₂ (x +1) – log₂ (x – 4) = 3

Log 2(x+1/x-4)

x+1/x-4 = 2³

x+1/x-4 = 8

x+1 = 8(x-4)

x+1 = 8x-32

x = 33/7.

Therefore, the solution to the problem Log₂ (x +1) – log₂ (x – 4) = 3 is x = 33/7.

Example5. Solve Log₆ (x + 4) + log₆ (x – 2) = log₆ (4x)

Solution. Log₆ (x + 4) + log₆ (x – 2) = log₆ (4x)

Log₆((x+4)(x-2)) = logy(4x)

(x+4)(x-2) =(4x)

x²+2x-8 = 4x

x²-2x-8 = 0

(x+2)(x-4) = 0

X=-2 or x = 4

X=4

Therefore, the solution to the problem Log₆ (x + 4) + log₆ (x – 2) = log₆ (4x) is x = 4.

Logarithm problem:

The discrete logarithm problem is described as finding the discrete logarithm to the base g of h in the group G given a group G, a generator g of the group, and an element h of G. Problems using discrete logarithms are not always difficult. Finding discrete logarithms is difficult depending on the groups.

Problem1. In logarithmic form, 5³ = 125.

Solution. 5³ = 125

As we know,

ab = c ⇒ logac=b

Therefore;

Log₅125 = 3

Problem 2. In exponential form, write log₁₀1 = 0.

Solution. Given, log₁₀1 = 0

Logac=b ⇒ ab = c

Hence,

10⁰ = 1

Problem3. Calculate the log of 32 to the fourth power.

Solution: log₄32 = x

4x = 32

(2²)x = 2x2x2x2x2

2²x = 2⁵

2x=5

x=5/2Therefore,

log432 =5/2.

Problem4. Find x if log5(x-7)=1.

Solution. Given,

Log5(x-7)=1

We can write; using logarithm rules.

5¹ = x-7

5 = x-7

X=5+7

X=12.

Problem5. Express log(75/16)-2log(5/9)+log(32/243) in terms of log 2 and log 3.

Solution. Log(75/16)-2log(5/9)+log(32/243) 

Since, nlogam=logamn

⇒log(75/16)-log(5/9)² +log(32/243)

⇒log(75/16)-log(25/81)+log(32/243)

Since, logam-logan=loga(m/n)

⇒log[(75/16)÷(25/81)]+log(32/243)

⇒log[(75/16)×(81/25)]+log(32/243)

⇒log(243/16)+log(32/243)

Since, logam+logan=logamn

⇒log(32/16)

⇒log2.

Conclusion:

The ability of logarithms to solve exponential equations accounts for much of their potency. Sound (decibel measurements), earthquakes (Richter scale), star brightness, and chemistry are all examples of this (pH balance, a measure of acidity and alkalinity).