Answer :
Sure! Let's solve the problem step by step.
The problem is to find [tex]\(\log_8(0.66)\)[/tex].
### Step 1: Use the Change-of-Base Theorem
The change-of-base theorem states:
[tex]\[ \log_b(a) = \frac{\log_k(a)}{\log_k(b)} \][/tex]
We can use any base [tex]\( k \)[/tex], but for simplicity, we'll use the natural logarithm (base e). So, in our case:
[tex]\[ \log_8(0.66) = \frac{\ln(0.66)}{\ln(8)} \][/tex]
### Step 2: Calculate the Natural Logarithms
We need to find the natural logarithms of 0.66 and 8:
[tex]\(\ln(0.66) \approx -0.4155\)[/tex]
[tex]\(\ln(8) \approx 2.0794\)[/tex]
### Step 3: Apply the Change-of-Base Formula
With the natural logarithms calculated, plug these values into the change-of-base formula:
[tex]\[ \log_8(0.66) = \frac{\ln(0.66)}{\ln(8)} = \frac{-0.4155}{2.0794} \][/tex]
### Step 4: Perform the Division
Now, divide the numbers:
[tex]\[ \frac{-0.4155}{2.0794} \approx -0.1998 \][/tex]
### Step 5: Round the Final Answer
Finally, rounding to four decimal places gives us:
[tex]\[ \log_8(0.66) \approx -0.1998 \][/tex]
So, the value of [tex]\(\log_8(0.66)\)[/tex] is approximately [tex]\(-0.1998\)[/tex].
The problem is to find [tex]\(\log_8(0.66)\)[/tex].
### Step 1: Use the Change-of-Base Theorem
The change-of-base theorem states:
[tex]\[ \log_b(a) = \frac{\log_k(a)}{\log_k(b)} \][/tex]
We can use any base [tex]\( k \)[/tex], but for simplicity, we'll use the natural logarithm (base e). So, in our case:
[tex]\[ \log_8(0.66) = \frac{\ln(0.66)}{\ln(8)} \][/tex]
### Step 2: Calculate the Natural Logarithms
We need to find the natural logarithms of 0.66 and 8:
[tex]\(\ln(0.66) \approx -0.4155\)[/tex]
[tex]\(\ln(8) \approx 2.0794\)[/tex]
### Step 3: Apply the Change-of-Base Formula
With the natural logarithms calculated, plug these values into the change-of-base formula:
[tex]\[ \log_8(0.66) = \frac{\ln(0.66)}{\ln(8)} = \frac{-0.4155}{2.0794} \][/tex]
### Step 4: Perform the Division
Now, divide the numbers:
[tex]\[ \frac{-0.4155}{2.0794} \approx -0.1998 \][/tex]
### Step 5: Round the Final Answer
Finally, rounding to four decimal places gives us:
[tex]\[ \log_8(0.66) \approx -0.1998 \][/tex]
So, the value of [tex]\(\log_8(0.66)\)[/tex] is approximately [tex]\(-0.1998\)[/tex].
