What is actually a logarithmic equation in algebra?
Any equation that is expressed in the form of logarithms and involving a variable as any of the values in either of the sides of the equals to sign can be classified as a logarithmic equation. The simplest example of this would be;
log2(x) = log2(30) ; x = 30
So, this was a type of equation with only 2 logs and both their bases being equal the arguments on both sides had to be equal for the equation to hold good.
Now, how to solve logarithmic equations?
Logarithmic equations can be solved by two methods other than punching stuff into a calculator and reaching the result straight away. One method is as shown above, known as solving from the definition. We will however see further examples with a higher difficulty level for the same. The next method is solving by exponentials. We will discuss it in detail individually later. Let’s see the first one first.
Solving from the definition:
Example: (for simplicity we’ll keep the base of the logs same)
2log10(x) = log10(4) + log10(x+1) [log10 is also denoted as ln]
2ln(x) = ln ((4)(x+1))
2ln(x) = ln (4x + 4) [ now we need to make it comparable on both sides]
ln (x)2 = ln (4x + 4) [by the log rule]
x2 = 4x + 4
x2 ~ 4x ~ 4 = 0
(x + 2)(x ~ 2) = 0
x = 2, ~2
Answer: x = 2, as we can’t have a logarithm of a negative integer.
Now, let’s move to the second method, i.e. :
Solving by logarithm:
This method involves the application of a relation, which is;
If y = bX , then logb(y) = X . Here we need to remember that whatever is the exponential will become the ‘equals to’ when you convert it to logarithmic form and vice versa. And the integer on which the exponential was raised will become the base of the log. This would help you solve any equation simply and faster. Now let’s take an example.
log2(x + 2) + log2(x – 2) = 3
log2 (x+2) (x – 2) = 3
Going by the above relation; (x + 2) (x – 2) = 23
x2 – 4x – 4 = 8
x2 – 4x – 4 – 8 = 0
x2 – 4x – 12 = 0
(x – 6) (x + 2) = 0
either x – 6 = 0 or x + 2 = 0 , If x – 6 = 0 then x = 6 ; & If x + 2 = 0 then x = – 2
Since we can’t have a negative integer or zero for logarithm arguments (x = – 2) stands invalid, hence x = 6 .
There can be several other ways of a logarithmic equation but however complex it be, you can always break it down to simpler forms with this method of solving and reach the correct solution.