write (27a^-3)^(-2/3) in simplest form.

solve log(6x+11)=2

please and thanks

I need to see your work on the first.

On the second, take the antilog of each side

what is the antilog?

I'm not quite sure what to do for the first, but this is what I tried to do:

(27a^-3)^(-2/3)
=the cubed root of (27a^-3)^-2
=the cubed root of 27^-2*a^6

To simplify the expression (27a^-3)^(-2/3), we can use the properties of exponents.

Step 1: Apply the power of a power rule.
(27a^-3)^(-2/3) = 27^(-2/3) * (a^-3)^(-2/3)

Step 2: Simplify the exponents separately.
27^(-2/3) = (3^3)^(-2/3) = 3^(-2) = 1/3^2 = 1/9

(a^-3)^(-2/3) = a^(-3 * -2/3) = a^2

Step 3: Combine the simplified exponents.
(27a^-3)^(-2/3) = 1/9 * a^2

Therefore, the simplified form of (27a^-3)^(-2/3) is 1/9 * a^2.

Now, moving on to the second question:

To solve the equation log(6x + 11) = 2, we need to eliminate the logarithm by taking the antilogarithm.

The antilogarithm, also known as the exponential function, is the inverse operation of a logarithm. It helps us remove the logarithm and find the value of the argument.

In this case, the base of the logarithm is not specified, so we assume it's a common logarithm with a base of 10. The antilogarithm of a logarithm with a base of 10 is the 10 raised to the power of the logarithm value.

Step 1: Rewrite the equation using the exponential notation.
10^(log(6x + 11)) = 10^2

Step 2: Simplify the equation.
6x + 11 = 100

Step 3: Solve for x.
6x = 100 - 11
6x = 89
x = 89/6

The solution to the equation log(6x + 11) = 2 is x = 89/6.

Please note that if the base of the logarithm is different, you would use the corresponding antilogarithm. For example, if the base is e (natural logarithm), you would use the exponential function e^x.