Review Indices law and Solving Exponential Equation

Apply $$\sqrt[n]{a^m}=a^{\frac{m}{n}}$$

$$\frac{1}{3\times3\times3\times3}=\frac{1}{3^4}$$

$$=3^{-4}$$

$$\frac{16a^3b^2}{8ab}=\frac{16}{8}\times\frac{a^3}{a}\times\frac{b^2}{b}$$

$$=2\times a^{\left(3-1\right)}\times b^{\left(2-1\right)}$$

$$=2a^2b$$

$$\frac{5a^5b^6}{25a^7b^3}=\frac{25}{5}\times\frac{a^5}{a^7}\times\frac{b^6}{b^3}$$

$$=\frac{1}{5}\times a^{\left(5-7\right)}\times b^{\left(6-3\right)}$$

$$=\frac{1}{5}\times a^{-2}\times b^3$$

$$=\frac{1}{5}\times\frac{1}{a^2}\times b^3$$

$$=\frac{b^3}{5a^2}$$

$$\frac{35\left(ab^2\right)^4}{5ab^2}=\frac{35a^4b^{\left(2\times4\right)}}{5ab^2}$$

$$=\frac{35}{5}\times\frac{a^4}{a}\times\frac{b^8}{b^2}$$

$$=7\times a^{\left(4-1\right)}\times b^{\left(8-2\right)}$$

$$=7a^3b^6$$

$$\frac{\left(a^{\frac{1}{2}}b^2\right)^6}{a^4b^2}=\frac{a^{\left(\frac{1}{2}\times6\right)}b^{\left(2\times6\right)}}{a^4b^2}=\frac{a^3b^{12}}{a^4b^2}$$

$$=a^{\left(3-4\right)}b^{\left(12-2\right)}$$

$$=a^{-1}b^{10}$$

$$=\frac{b^{10}}{a}$$

$$3^x=27$$

$$3^x=3^3$$

So, $$x=3$$

Express both sides with the same base

$$2^x=144$$

$$2^x=2^7$$

So $$x=7$$

Express 81 in the base of 3.

(Hint: use factor tree)

$$3^{x+1}=81$$

$$3^{x+1}=3^4$$

Then solve the index equation

$$x+1=4$$

$$x=4-1=3$$