Question

If f (x) = 16*+log2 x Find f (1/2)​

Answer 1

$\displaystyle \sf{If \: \: f(x) = {16}^{x} + log_{2} x \: \: then \: \: f \bigg( \frac{1}{2} \bigg) }= 3$

Correct question :

$\displaystyle \sf{If \: \: f(x) = {16}^{x} + log_{2} x \: \: find \: \: f \bigg( \frac{1}{2} \bigg) }$

Given :

$\displaystyle \sf{f(x) = {16}^{x} + log_{2} x }$

To find :

$\displaystyle \sf{f \bigg( \frac{1}{2} \bigg) }$

Formula :

$\displaystyle \sf{1. \: \: log \bigg( \frac{a}{b} \bigg) = log(a) - log(b) }$

$\sf{2. \: \: log_{a}(a) = 1}$

Solution :

Step 1 of 2 :

Write down the given function

Here the given function is

$\displaystyle \sf{ f(x) = {16}^{x} + log_{2} x }$

Step 2 of 2 :

Find the required value

$\displaystyle \sf{ f(x) = {16}^{x} + log_{2} x }$

Putting x = 1/2 we get

$\displaystyle \sf{f \bigg( \frac{1}{2} \bigg) = {16}^{ \frac{1}{2} } + log_{2} \bigg( \frac{1}{2} \bigg ) }$

$\displaystyle \sf\implies f \bigg( \frac{1}{2} \bigg) = {( {4}^{2} )}^{ \frac{1}{2} } +\bigg[ log_{2}(1) - log_{2}(2) \bigg] \: \: \: \bigg[ \: \because \:log \bigg( \frac{a}{b} \bigg) = log(a) - log(b) \bigg]$

$\displaystyle \sf\implies f \bigg( \frac{1}{2} \bigg) = {( {4}^{} )}^{ (2 \times \frac{1}{2}) } +\bigg(0 - 1 \bigg)\: \: \: \bigg[ \: \because \: log_{a}(a) = 1 \bigg]$

$\displaystyle \sf\implies f \bigg( \frac{1}{2} \bigg) = {4}^{1} - 1$

$\displaystyle \sf\implies f \bigg( \frac{1}{2} \bigg) = 4 - 1$

$\displaystyle \sf\implies f \bigg( \frac{1}{2} \bigg) = 3$

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