Question

Solve
$log_{2}( log_{2}x) = 4$
​

Answer 1

$\large\underline{\sf{Solution-}}$

Given Logarithmic equation is

$\rm \: log_{2}( log_{2}x) = 4$

We know,

$\boxed{ \rm{ \: log_{a}(b) = c \: \rm\implies \:b = {a}^{c} \: \: }}$

So, using this, we get

$\rm \: log_{2}(x) = {2}^{4}$

$\rm \: log_{2}(x) = 16$

Again we know,

$\boxed{ \rm{ \: log_{a}(b) = c \: \rm\implies \:b = {a}^{c} \: \: }}$

So, using this result, we get

$\rm\implies \:\boxed{ \rm{ \:\rm \: x = {2}^{16} \: \: }}$

$\rule{190pt}{2pt}$

Additional Information :-

$\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{ \red{More \: Formulae}}}} \\ \\ \bigstar \: \bf{ log_{x}(x) = 1}\\ \\ \bigstar \: \bf{ log_{x}( {x}^{y} ) = y}\\ \\ \bigstar \: \bf{ log_{ {x}^{z} }( {x}^{w} ) = \dfrac{w}{z} }\\ \\ \bigstar \: \bf{ log_{a}(b) = \dfrac{logb}{loga} }\\ \\ \bigstar \: \bf{ {e}^{logx} = x}\\ \\ \bigstar \: \bf{ {e}^{ylogx} = {x}^{y}}\\ \\ \bigstar \: \bf{log1 = 0}\\ \end{array} }}\end{gathered}\end{gathered}\end{gathered}$