# Solving logarithmic functions using Logarithmic Identities

**Example 1: Solve ***log*_{2}(x) + log_{2}(x – 2) = 3

__Solution__: Here we need to use logarithmic identities to combine the two terms on the left-hand side of the equation:

*log*_{2}(x) + log_{2}(x – 2) = 3

*log*_{2} ((x) (x – 2)) = 3

*log*_{2}(x^{2} – 2x) = 3

Then we’ll use the relationship to convert the log form to the corresponding exponential form, and then move further,

*log*_{2}(x^{2} – 2x) = 3

*2*^{3} = x^{2} – 2x* *

*8 = *x^{2} – 2x* *

*0 = *x^{2} – 2x* **– 8 *

*0 = (*x* **– 4) (*x* **+ 2) *

x* **= 4, –2*

But if *x = –2*, then "*log*_{2}(x)", from the original logarithmic equation, will have a negative number for its argument (as will the term "*log*_{2}(x – 2)"). Since logs cannot have zero or negative arguments, then the solution to the original equation cannot be *x = –2*.

The solution is x* **= 4*.

**
Example 2: Solve ***log*_{2}(x^{2}) = (log_{2}(x))^{ }^{2}.

__Solution__: First, we'll write out the square on the right-hand side:

*log*_{2}(x^{2}) = (log_{2}(x))^{ }^{2}

*log*_{2}(x^{2}) = (log_{2}(x)) (log_{2}(x))

Then we’ll apply the log rule to move the "squared", from inside the log on the left-hand side of the equation, out in front of that log as a multiplier. Then we'll move that term to the right-hand side:

*2log*_{2}(x) = [log_{2}(x)] [log_{2}(x)]

*0 = [log*_{2}(x)] [log_{2}(x)] – 2log2(x)

Factorize the above equation to obtain

*0 = [log*_{2}(x)] [log_{2}(x) – 2]

*log*_{2}(x) = 0 or log2(x) – 2 = 0

*2*^{0} = x or log2(x) = 2

*1 = x or 22 = x *

*1 = x or 4 = x*

**
Example 3: Solve **

__Solution:__ Since *4/x* is the base of a log function x is greater than 0.

The equation is then equivalent with *(4/x)*^{ }^{2} = x^{2}- 6

<=> *16 = x*^{4}- 6x^{2}^{ }

Now, let *x*^{2} = t

<=> *16 = t*^{2}- 6t

<=> *t=-2 or t=8*

<=> *x = sqrt (8) *(since x >0)

The solution is *x = 1, 4.*