Evaluate the expression \[{{2}^{{{\log }_{2}}45}}?.\]

Hint: In the given question we have to evaluates the value of \[{{2}^{{{\log }_{2}}45}}.\] For this simplify the expression by using logarithmic properties. Apply the required property to get the solution of the given expression.

Complete step by step solution:
In this question, the expression for evaluation is \[{{2}^{{{\log }_{2}}45}}.\]
Consider the value of \[x\]is \[{{2}^{{{\log }_{2}}45}}.\]
Then,
\[\Rightarrow \]\[x={{2}^{{{\log }_{2}}45}}\]
Now, taking \[\log \]with base \[2\]on both sides of equation
Therefore above expression can be written as,
\[\Rightarrow {{\log }_{2}}x={{\log }_{2}}{{2}^{\left( {{\log }_{2}}45 \right)}}\]
Now, we know that the property of logarithm which is \[\log {{(a)}^{b}}\]we can write this as \[b\log a\] i.e. \[\log {{a}^{b}}=b\log a.\]
Therefore, by using property the above expression will be written as,
\[\Rightarrow {{\log }_{2}}x=\left( {{\log }_{2}}45 \right){{\log }_{2}}2\]
Now, using property of \[{{\log }_{a}}b=\dfrac{\log b}{\log a}\]above expression can be written as,
\[\Rightarrow \dfrac{\log x}{\log 2}=\left( {{\log }_{2}}45 \right){{\log }_{2}}2\]
Also, \[{{\log }_{2}}45=\dfrac{\log 45}{\log 2}\] put this value in above expression
We have,
\[\Rightarrow \dfrac{\log x}{\log 2}=\dfrac{\log 45}{\log 2}{{\log }_{2}}2\]
Now, multiply with \[\log 2\] to

Now, taking log with base \[2\] on both sides of the equation.
Therefore above expression can be written as,
\[\Rightarrow \]\[{{\log }_{2}}x={{\log }_{2}}{{2}^{\left( {{\log }_{2}}45 \right)}}\]
Now, we know that the property of logarithm which is \[\log {{\left( a \right)}^{b}}\] we can write this as \[b \operatorname{loga}\] i.e. \[\log {{a}^{b}}=b\log a\]
Therefore, by using property the above expression will be written as,
\[\Rightarrow {{\log }_{2}}x=\left( {{\log }_{2}}45 \right){{\log }_{2}}2\]
Now, using property of \[{{\log }_{a}}b=\dfrac{\operatorname{logb}}{\log a}\] above expression can be written as,
\[\Rightarrow \]\[\dfrac{\log x}{\log 2}=\left( {{\log }_{2}}45 \right){{\log }_{2}}2\]
Also, \[{{\log }_{2}}45=\dfrac{\log 45}{\log 2}\] put this value in above expression
We have,
\[\Rightarrow \]\[\dfrac{\log x}{\log 2}=\dfrac{\log 45}{\log 2}{{\log }_{2}}2\]
Now, multiply with \[\log 2\] to both sides of the equation. We have
\[\Rightarrow \]\[\dfrac{\log x}{\log 2}\times \log 2=\dfrac{\log 45}{\log 2}\times \log 2(lo{{g}_{2}}2)\]
Now, cancelling common factor above expression can be written as,
\[\Rightarrow log x=\log 45.{{\log }_{2}}2\]
Then, \[{{\log }_{2}}2=\dfrac{\log 2}{\log 2}\] put this value in above equation \[\operatorname{log x}=log45.\dfrac{\log 2}{\log 2}\]
By cancelling the common factor we have the modified equation.
\[\Rightarrow log x=\log 45\]
Now we know that if \[\log a=\log b\] then \[a=b\]
Therefore,
\[\Rightarrow x=45\]
Hence, the evaluated value of the expression \[{{2}^{{{\log }_{2}}45}}\] is \[45\].

Note: The logarithm is the inverse function to exponentiation. That means the logarithm of a given number \[x\] is the exponent to which another fixed number the base \[b,\] must be raised to produce their number \[x.\] For example, the base ten logarithm of \[100\] is \[2\] because ten raised to the power of two is \[100\] that is \[\log 100=2\] because \[{{10}^{2}}=100.\]