Answer :
Let's solve for [tex]\(a_3\)[/tex] step by step using the given recurrence relation [tex]\( a_n = a_{n-1} + \log_{n+1}(n-1) \)[/tex] and the initial condition [tex]\( a_1 = 8 \)[/tex].
### Step 1: Calculate [tex]\( a_2 \)[/tex]
Given:
[tex]\[ a_1 = 8 \][/tex]
The recurrence relation for [tex]\( a_2 \)[/tex] is:
[tex]\[ a_2 = a_1 + \log_3(1) \][/tex]
Here, we need to evaluate the logarithm:
[tex]\[ \log_3(1) = 0 \][/tex]
since any number to the power of 0 is 1.
Thus:
[tex]\[ a_2 = 8 + 0 = 8 \][/tex]
### Step 2: Calculate [tex]\( a_3 \)[/tex]
Now, use the value of [tex]\( a_2 \)[/tex] to find [tex]\( a_3 \)[/tex]:
[tex]\[ a_3 = a_2 + \log_4(2) \][/tex]
Next, we need to evaluate this logarithm:
[tex]\[ \log_4(2) = \frac{\log_2(2)}{\log_2(4)} \][/tex]
[tex]\[ \log_2(2) = 1 \][/tex]
[tex]\[ \log_2(4) = 2 \][/tex]
So:
[tex]\[ \log_4(2) = \frac{1}{2} \][/tex]
Thus:
[tex]\[ a_3 = 8 + \frac{1}{2} = 8.5 \][/tex]
Therefore, the value of [tex]\( a_3 \)[/tex] is:
[tex]\[ \boxed{8.5} \][/tex]
### Step 1: Calculate [tex]\( a_2 \)[/tex]
Given:
[tex]\[ a_1 = 8 \][/tex]
The recurrence relation for [tex]\( a_2 \)[/tex] is:
[tex]\[ a_2 = a_1 + \log_3(1) \][/tex]
Here, we need to evaluate the logarithm:
[tex]\[ \log_3(1) = 0 \][/tex]
since any number to the power of 0 is 1.
Thus:
[tex]\[ a_2 = 8 + 0 = 8 \][/tex]
### Step 2: Calculate [tex]\( a_3 \)[/tex]
Now, use the value of [tex]\( a_2 \)[/tex] to find [tex]\( a_3 \)[/tex]:
[tex]\[ a_3 = a_2 + \log_4(2) \][/tex]
Next, we need to evaluate this logarithm:
[tex]\[ \log_4(2) = \frac{\log_2(2)}{\log_2(4)} \][/tex]
[tex]\[ \log_2(2) = 1 \][/tex]
[tex]\[ \log_2(4) = 2 \][/tex]
So:
[tex]\[ \log_4(2) = \frac{1}{2} \][/tex]
Thus:
[tex]\[ a_3 = 8 + \frac{1}{2} = 8.5 \][/tex]
Therefore, the value of [tex]\( a_3 \)[/tex] is:
[tex]\[ \boxed{8.5} \][/tex]
