Multiplying like bases means add the exponents, so x+4 = 12, or x = 8.

Raising a power to a power means multiply the exponents, so 3y = 15, or y = 5.

x - y = 8 - 5 = 3.

The first step is to express both sides of the equation with equal bases, in this case 3. The equation becomes 3^3p = 3^2q. So then 3p = 2q, and q = (3/2)p is our answer. 

27^2/3 is 27 squared and cube-rooted. We want to pick the easier operation first. Here that is the cube root. To see that, try both operations. 

27^2/3 = (27²)^1/3 = 729^1/3 OR

27^2/3 = (27^1/3)² = 3²

Obviously 3² is much easier. Either 3² or 729^1/3 will give us the correct answer of 9, but with 3² it is readily apparent. 

To solve this problem, we will have to take the log of both sides to bring down our exponents. By doing this, we will get \(\displaystyle \dpi{100} \small a\ast log\left (\frac{1}{3} \right )= b\ast log\left ( 27 \right )\).

To solve for \(\displaystyle \dpi{100} \small \frac{a}{b}\) we will have to divide both sides of our equation by \(\displaystyle \dpi{100} \small log\frac{1}{3}\) to get \(\displaystyle \dpi{100} \small \frac{a}{b}=\frac{log\left ( 27 \right )}{log\left ( \frac{1}{3} \right )}\).

\(\displaystyle \dpi{100} \small \frac{log\left ( 27 \right )}{log\left ( \frac{1}{3} \right )}\) will give you the answer of –3.

We use two properties of logarithms: 

\(\displaystyle log(xy) = log (x) + log (y)\)

\(\displaystyle log(x^{n}) = nlog (x)\)

So \(\displaystyle \log 12=2 \log2+\log3\)

\(\displaystyle x^{m}\ast x^{n} = x^{m + n}\), here \(\displaystyle m=-3\) and \(\displaystyle n=6\), hence \(\displaystyle -3+6=3\).

\(\displaystyle \left ( \frac{2}{3} \right )^{x+1} = \frac{27}{8}\) = \(\displaystyle \left ( \frac{3}{2} \right )^{3} = \left ( \frac{2}{3} \right )^{-3}\)

 \(\displaystyle x+1=-3\) which means \(\displaystyle x=-4\)

Remember the laws of exponents. In particular, when the base is nonzero:

\(\displaystyle b^{m+n} = b^m b^n \qquad (b^m)^n = b^{mn} \\ (b \times c)^n = b^n c^n \qquad b^{-n} = 1/b^n\)

An effective way to compare these statements, is to convert them all into exponents with base 2. The original statement becomes:

\(\displaystyle 4^{x+2} / 2^y = 4^{x+2} (2^{-y}) = (2^2)^{x+2} 2^{-y} = 2^{2(x+2) } 2^{-y} = 2^{2x+4-y}\)

This is identical to statement I. Now consider statement II:

\(\displaystyle 2^{x+2} / 4^y = 2^{x+2} ( 4^{-y} ) = 2^{x+2} ( 2^2 )^{-y} = 2^{x+2} 2^{-2y} = 2^{x+2-2y} \neq 2^{2x+4-y}\)

Therefore, statement II is not identical to the original statement. Finally, consider statement III:

\(\displaystyle (4^x)(4^2 ) (2^{-y}) = (2^2)^x (2^2)^2 (2^{-y}) = (2^{2x} )( 2^4 ) (2^{-y}) = 2^{2x+4-y}\)

which is also identical to the original statement. As a result, only I and III are the same as the original statement. 

Using properties of radicals e.g., \(\displaystyle \sqrt[m]{x^{n}} = x^{\frac{n}{m}}\)

we get \(\displaystyle \sqrt[3]{\left ( x-1 \right )^{2}} = \left ( x-1 \right )^{\frac{2}{3}}\)

Properties of Radicals

\(\displaystyle \sqrt[]{x} = x^{\frac{1}{2}}\)

\(\displaystyle \sqrt[3]{x} = x^{\frac{1}{3}}\)

\(\displaystyle x^{\frac{5}{3}}\)