Hello, I need some assistance with this precalculus homework question, please?HW Q3

EXPLANATION:
Given:
We are given the following information;
[tex]2^{-x}=4.5[/tex]Required:
We are required to find the value of x (that is, the solution to this equation).
Step-by-step solution;
To do this we would re-write this equation by applying the law of exponents, which is as follows;
[tex]\begin{gathered} If: \\ f(x)=g(x) \\ Then: \\ ln(f(x))=ln(g(x)) \end{gathered}[/tex]With this, we will take the natural log of both sides of the equation;
[tex]ln(2^{-x})=ln(4.5)[/tex]Next we, take the left side of the equation and apply the law of logs, as shown below;
[tex]log_ba^x=xlog_ba[/tex]Therefore, we can refine the left side;
[tex]ln(2^{-x})=-xln2[/tex]We can now re-write the entire equation as shown below;
[tex]-xln2=ln4.5[/tex]Divide both sides of the equation by ln(2);
[tex]\frac{-xln2}{ln2}=\frac{ln4.5}{ln2}[/tex][tex]-x=\frac{ln(4.5)}{ln(2)}[/tex]Multiply both sides of the equation by negative 1;
[tex]x=-\frac{ln(4.5)}{ln(2)}[/tex]We now have the exact answer for x.
To solve for the value of x rounded to 3 decimal places;
[tex]x=-\frac{ln(4.5)}{ln(2)}[/tex]With the use of a calculator, we would now have;
[tex]x=-\frac{1.504077}{0.693147}[/tex][tex]x=-2.16992[/tex]We can round this to 3 decimal places and we'll have;
[tex]x=-2.169[/tex]Therefore;
ANSWER:
[tex]\begin{gathered} (1) \\ A:The\text{ }solution\text{ }set\text{ }is:x=-\frac{ln(4.5)}{ln(2)} \end{gathered}[/tex][tex]\begin{gathered} (2) \\ A:The\text{ }solution\text{ }set\text{ }is:x=-2.169 \end{gathered}[/tex]