2.6 Other types of equations

Whether we take the root first or the power first depends on the number. It is easy to find the cube root of 8, so rewrite $8^{\frac{2}{3}}$ as ${\left(8^{\frac{1}{3}}\right)}^2.$

The way to remove the exponent on x is by raising both sides of the equation to a power that is the reciprocal of $\frac{5}{4},$ which is $\frac{4}{5}.$

This equation involves rational exponents as well as factoring rational exponents. Let us take this one step at a time. First, put the variable terms on one side of the equal sign and set the equation equal to zero.

Now, it looks like we should factor the left side, but what do we factor out? We can always factor the term with the lowest exponent. Rewrite $x^{\frac{1}{2}}$ as $x^{\frac{2}{4}}.$ Then, factor out $x^{\frac{2}{4}}$ from both terms on the left.

Where did $x^{\frac{1}{4}}$ come from? Remember, when we multiply two numbers with the same base, we add the exponents. Therefore, if we multiply $x^{\frac{2}{4}}$ back in using the distributive property, we get the expression we had before the factoring, which is what should happen. We need an exponent such that when added to $\frac{2}{4}$ equals $\frac{3}{4}.$ Thus, the exponent on x in the parentheses is $\frac{1}{4}.$

Let us continue. Now we have two factors and can use the zero factor theorem.

The two solutions are $x=0,$ $x=\frac{1}{81}.$