Okay, so I need to find the product of 6 base 8 multiplied by 7 base 8, and express the answer in base 8. Hmm, let's break this down step by step. First, I remember that when multiplying numbers in a different base, it might be easier to convert them to decimal (base 10), do the multiplication there, and then convert back to base 8. But maybe there's a way to do it directly in base 8 without converting? Let me think about both approaches.

Starting with the conversion method: 6 in base 8 is equal to 6 in decimal, right? Because 6 is less than 8. Similarly, 7 in base 8 is 7 in decimal. So multiplying 6 and 7 in decimal gives 42. Now, I need to convert 42 back to base 8. To do that, I divide 42 by 8. 8 times 5 is 40, so the quotient is 5 and the remainder is 2. Therefore, 42 in decimal is 52 in base 8. So the answer should be 52 base 8. Let me check if that's correct.

Alternatively, maybe I can try multiplying directly in base 8. Let's see how that would work. Multiplying 6_8 by 7_8. In base 8, the digits go from 0 to 7, so when multiplying, if the result is 8 or more, I need to carry over to the next place. Let's set it up like multiplication:

      6_8
    x 7_8
    -------
      ??

Starting from the rightmost digit: 6 times 7. In decimal, that's 42. Now, converting 42 to base 8: 8*5=40, remainder 2. So that's 5*8 + 2, which is 52_8. Wait, but we're in base 8 multiplication. So when multiplying 6_8 by 7_8, the product is 2 in the units place and carryover 5 (since 5*8=40). But since there are no more digits to multiply, we just write the carryover. So the result is 52_8. That's the same as the decimal method. So both methods give the same answer. That seems right.

But let me verify again. Let's do the multiplication step by step in base 8. 6 times 7. In base 10, that's 42. 42 divided by 8 is 5 with remainder 2. So the units place is 2 and carryover 5. Since there are no more digits to multiply, we just write the carryover in the next higher place. So that's 52_8. Yep, that's the same as before. So the answer should be 52 base 8.

Wait, but maybe I should check with another example to make sure I'm not making a mistake. Let's take a simpler multiplication, like 3_8 times 2_8. In decimal, that's 3*2=6, which is 6_8. So 3*2 in base 8 is 6. If I did it directly, 3*2=6, which is less than 8, so no carryover. Correct. Another example: 5_8 times 6_8. In decimal, 5*6=30. 30 divided by 8 is 3 with remainder 6. So 36_8. Let's check: 5*6=30. 30 in base 8 is 3*8 +6=36_8. Correct. So the method works.

Therefore, applying the same logic, 6_8 times 7_8 is 42 in decimal, which is 5*8 +2=52_8. So the answer is 52 base 8. I think that's solid. Let me just confirm once more. If I convert both numbers to decimal, multiply, and convert back, that's 6*7=42, 42 decimal is 52 octal. So yes, that's correct.

Alternatively, if I were to do the multiplication directly in base 8, even though I didn't need to carry over here, let's see. 6 times 7. 6*7=42. Then 42 divided by 8 is 5 with remainder 2. So in base 8, that's 52. So again, same result. Therefore, the answer is 52 in base 8. I think that's all. No mistakes detected in the process. So the final answer is 52_8.
Therefore, the product of $6_8 \cdot 7_8$ in base $8$ is $\boxed{52_8}$.
</think>