Evaluate the logarithm.(Can you give me a detailed step-by-step process of explaining how you evaluated. Also how a log works in general)

Given:
[tex]log_2(512)[/tex]To solve it, follow the steps below.
Step 01: Use the given hint (512 = 32*16).
[tex]log_2(32*16)[/tex]Step 02: Use the product rule for logarithms.
According to the product rule:
[tex]log(a*b)=loga+logb[/tex]Then,
[tex]log_2(32*16)=log_232+log_216[/tex]Step 03: Use the definition of the log to solve the problem.
Given the definition:
[tex]log_ba=c\Leftrightarrow b^c=a[/tex]So, let's solve each part of the equation.
[tex]log_232=x\Leftrightarrow2^x=32[/tex]In order to find x, let's factor 32.
32 | 2
16 | 2
8 | 2
4 | 2
2 | 2
1
2⁵ = 32
[tex]\begin{gathered} 2^x=2^5 \\ Then, \\ x=5 \end{gathered}[/tex]And,
[tex]log_232=5[/tex]Now, let's solve the second term.
[tex]\begin{gathered} log_216=y\Leftrightarrow2^y=16 \\ \end{gathered}[/tex]Let's factor 16:
16 | 2
8 | 2
4 | 2
2 | 2
1
2⁴ = 16
Substituting it in the equation:
[tex]\begin{gathered} 2^y=2^4 \\ Since\text{ }the\text{ }bases\text{ }are\text{ }the\text{ }same,\text{ }the\text{ }exponents\text{ }must\text{ }be\text{ }the\text{ }same\text{ }too \\ y=4 \end{gathered}[/tex]Then,
[tex]log_216=4[/tex]Step 04: Substitute the solutions and solve the equations.
[tex]\begin{gathered} log_232=5 \\ log_216=4 \\ Then, \\ log_232+log_216=5+4 \\ =9 \end{gathered}[/tex]Answer: 9.