Question

If $$f\left( x\right) =16^{x}+\log_{4} x$$, find $$f\left( \dfrac{1}{2} \right) ,f\left( \dfrac{1}{4} \right) $$

Answer 1

Hint: In this question it is given that $$f\left( x\right) =16^{x}+\log_{4} x$$, so we have to find the value of $$f\left( \dfrac{1}{2} \right) ,f\left( \dfrac{1}{4} \right) $$. Since the given function is the function of x, so to find the solution we have to put $$x=\dfrac{1}{2} ,\ x=\dfrac{1}{4}$$ in the given function.

Complete step-by-step answer:
Given function,
$$f\left( x\right) =16^{x}+\log_{4} x$$...............(1)
So in order to find $$f\left( \dfrac{1}{2} \right)$$ we have to put $$x=\dfrac{1}{2} $$ in the above function,
So when $$x=\dfrac{1}{2} $$,
$$f\left( \dfrac{1}{2} \right) =\left( 16\right)^{\dfrac{1}{2} } +\log_{4} \left( \dfrac{1}{2} \right) $$
$$=\left( 2^{4}\right)^{\dfrac{1}{2} } +\log_{4} 2^{-1}$$ [$$\because \dfrac{1}{a} =a^{-1}$$]
Now as we know that $$\left( a^{m}\right)^{n} =a^{m\times n}$$ and $$\log_{k} a^{b}=a\log_{k} b$$,
So by using these formula we can write the above equation as,
$$f\left( \dfrac{1}{2} \right) =2^{\left( 4\times \dfrac{1}{2} \right) }+\left( -1\right) \log_{4} 2$$
$$=2^{2}-\log_{4} 2$$
$$=2^{2}-\dfrac{1}{2} \times 2\log_{4} 2$$ [multiply and divide 2 in the second term]
$$=4-\dfrac{1}{2} \log_{4} 2^{2}$$ [since, $$a\log_{k} b=\log_{k} a^{b}$$]
$$=4-\dfrac{1}{2} \log_{4} 4$$
Now as we know that $$\log_{a} a=1$$, so therefore we can write,
$$f\left( x\right) =4-\dfrac{1}{2} \times 1$$
$$=4-\dfrac{1}{2}$$
$$=\dfrac{4\times 2-1}{2}$$
$$=\dfrac{7}{2}$$
Therefore, $$f\left( \dfrac{1}{2} \right) =\dfrac{7}{2}$$
Now when $$x=\dfrac{1}{4}$$
$$f\left( \dfrac{1}{4} \right) =\left( 16\right)^{\dfrac{1}{4} } +\log_{4} \left( \dfrac{1}{4} \right) $$
$$=\left( 2^{4}\right)^{\dfrac{1}{4} } +\log_{4} 4^{-1}$$
$$=2^{\left( 4\times \dfrac{1}{4} \right) }+\log_{4} 4^{-1}$$ [since, $$\left( a^{m}\right)^{n} =a^{m\times n}$$]
$$=2^{1}+\left( -1\right) \log_{4} 4$$ [since, $$\log_{k} a^{b}=a\log_{k} b$$]
$$=2-1$$ [since, $$\log_{a} a=1$$]
$$= 1$$
Therefore $$f\left( \dfrac{1}{4} \right) =1$$.

Note: While solving this type of question you need to know that if you plot this given function on graph paper then it will give us a curve where x is the independent variable and f(x) is dependent variable and when you put any value of x in that function then this will give you one output i.e, y coordinate. For example- in the given question we have put $$x=\dfrac{1}{4}$$ which give us y=f(x)=1, therefore $$\left( \dfrac{1}{4} ,1\right) $$ is the coordinate of a particular point on the given curve $$f\left( x\right) =16^{x}+\log_{4} x$$.