Question

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

Answer 1

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 ${{10}^{2}}=100.$