If $log_{2}\,6+\frac{1}{2x} = log_{2}\left(2^{\frac{1}{x}} + 8\right)$, then the values of x are
- $\frac{1}{4}, \frac{1}{3}$
- $\frac{1}{4}, \frac{1}{2}$
- $-\frac{1}{4}, \frac{1}{2}$
- $\frac{1}{3}, \frac{1}{^{-}2}$
The Correct Option is B
Solution and Explanation
$\Rightarrow a^{2}-6a+8 = 0 \,\Rightarrow\,a = 2\,\Rightarrow\,x = \frac{1}{4}, \frac{1}{2}$
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Concepts Used:
Functions
A function is a relation between a set of inputs and a set of permissible outputs with the property that each input is related to exactly one output. Let A & B be any two non-empty sets, mapping from A to B will be a function only when every element in set A has one end only one image in set B.
Kinds of Functions
The different types of functions are -
One to One Function: When elements of set A have a separate component of set B, we can determine that it is a one-to-one function. Besides, you can also call it injective.
Many to One Function: As the name suggests, here more than two elements in set A are mapped with one element in set B.
Moreover, if it happens that all the elements in set B have pre-images in set A, it is called an onto function or surjective function.
Also, if a function is both one-to-one and onto function, it is known as a bijective. This means, that all the elements of A are mapped with separate elements in B, and A holds a pre-image of elements of B.
Read More: Relations and Functions