Question

1)log₃ + log₃6= 2) 81^1/4 - 3√3 × 3^1/2=

Answer 1

Решение дано на фото.

Answer 2

It seems like there might be some typos or missing information in your equations. Let's try to correct and solve them step-by-step:

1. Assuming the first equation is: log₃(x) + log₃(6) = ?

To solve this equation, we can use the logarithmic rule that states: logₐ(b) + logₐ(c) = logₐ(bc).

So, in this case, we have: log₃(x) + log₃(6) = log₃(x * 6)

Now, we want to simplify the right side to get a single logarithmic term. Since 6 is equal to 3 * 2, we can rewrite it as:

log₃(x * 6) = log₃(x * (3 * 2)) = log₃(3x * 2)

Finally, we can use the logarithmic rule again:

log₃(3x * 2) = log₃(3x) + log₃(2)

So, the simplified equation is: log₃(x) + log₃(6) = log₃(3x) + log₃(2)

2. Assuming the second equation is: 81^(1/4) - 3√3 × 3^(1/2) = ?

Let's solve this step-by-step:

81^(1/4) is the fourth root of 81, which is 3 because 3^4 = 81.

3√3 is the cube root of 3, which remains as 3.

3^(1/2) is the square root of 3, which is approximately 1.732.

Now, we can substitute these values into the equation:

81^(1/4) - 3√3 × 3^(1/2) = 3 - (3 × 1.732) = 3 - 5.196 ≈ -2.196

So, the result of the second equation is approximately -2.196.