How do you solve the algebraic expression $2{{e}^{0.5x}}=45$ ?

Hint: We are given an expression in which an exponential term is being equated to a constant term, 45. In order to find the solution of the given equation, we must find the value of x. Thus, we shall first make the coefficient of the exponential term equal to 1 by dividing the entire equation by 2. Further we will take logarithms on the left hand side and the right hand side of the equation and simplify the equation more.

Complete step by step solution:
Given that $2{{e}^{0.5x}}=45$.
In order to simplify the present equation, we shall divide it by 2 on the left hand side as well as the right hand side and make the coefficient of the exponential term equal to 1.
$\Rightarrow {{e}^{0.5x}}=\dfrac{45}{2}$
Now we shall take the logarithm with base e on both sides of the equation.
$\Rightarrow \log {{e}^{0.5x}}=\log \dfrac{45}{2}$
Here, we will use the property of logarithm, $\log {{a}^{b}}=b\log a$.
 $\Rightarrow 0.5x\log e=\log \dfrac{45}{2}$
We know that ${{\log }_{e}}e=1$, thus substituting this value we get,
 \[\Rightarrow \left( 0.5x \right)1=\log \dfrac{45}{2}\]
Also, using the logarithmic property, $\log \dfrac{a}{b}=\log a-\log b$, we get
\[\Rightarrow 0.5x=\log 45-\log 2\]
We know that $9\times 5=45$,
\[\Rightarrow 0.5x=\log \left( 9 \right)\left( 5 \right)-\log 2\]
Therefore, we can use the logarithmic property $\log ab=\log a+\log b$ here.
\[\Rightarrow 0.5x=\log 9+\log 5-\log 2\]
\[\Rightarrow x=\left( \dfrac{\log 9+\log 5-\log 2}{0.5} \right)\]
\[\Rightarrow x=2\left( \log 9+\log 5-\log 2 \right)\]
Since $\log 2=0.693,\log 5=1.60$ and $\log 9=2.19$ , hence substituting this value, we get
$\Rightarrow x=2\left( 2.19+1.60-0.693 \right)$
$\Rightarrow x=2\left( 3.097 \right)$
$\Rightarrow x=6.194$
Therefore, the solution for the given exponential equation $2{{e}^{0.5x}}=45$ is\[x=2\left( \log 9+\log 5-\log 2 \right)\] or $x=6.194$.

Note: Another way by which we could have simplified the term log9 and found out its value is by using the logarithmic property $\log {{a}^{b}}=b\log a$ as log9 can also be written as $\log {{3}^{3}}$. Thus, $\log 9=3\log 3$ and we could have substituted the value of log 3 to find the value of log 9.