to view answer.Sometimes we may need to solve equations that involve an expression that is raised to a power. For instance, we might have an equation with a fractional exponent, such as:
$$3\left( {x - 2} \right)^{\Large\frac{1}{5}} = 9$$
In these cases, remember what it means to “solve an equation for x”. It means to find the value or values of x that make the equation true. The way we do this is by changing the form of the equation, while making sure to preserve the equality (in other words, to keep the equation “balanced” so that the two sides remain equal). In general this means performing operations to both sides of the equation until it becomes obvious what x must be – which is generally when x is alone on one side of the equation.
So, let’s look at this example:
Solve for x. That is, find the values of x that make the equation true.
$$
3\left( {x - 2} \right)^{{\large\frac{1}{5}}} = 9
$$
1. First, we note that on the left hand side of the equation, we have the quantity $$\left( {x - 2} \right)^{\large{\frac{1}{5}}} $$ multiplied by 3. Thus, we can divide both sides of the equation by 3 to get:
$${\Large\frac{{3\left( {x - 2} \right)^{\frac{1}{5}} }}{3}} ={\Large \frac{9}{3}}$$
which is the same as:
$$\left( {x - 2} \right)^{\large{\frac{1}{5}}} = 3$$
2. Now we have a quantity $$\left( {x - 2} \right)$$ raised to the power of $${\Large\frac{1}{5}}$$ on the left hand side. By raising each side to the power of 5 the exponent on the left hand side becomes 1:
$$\left( {\left( {x - 2} \right)^{\large{\frac{1}{5}} }} \right)^5 = \left( 3 \right)^5 $$
And so, we have:
$$x - 2 = 3^5 $$
Or:
$$x - 2 = 243$$
3. Finally, adding 2 to each side, we get:
$$x = 245$$
So, the value of x that makes the original equation true is 245.
Solve for z. That is, find the values of z that make the equation true.
$$\left( {z + 3} \right)^{\large{\frac{2}{3}} } = 5^{\large{\frac{1}{3}}} $$
1. This time, both sides of the equation have fractional exponents. Don’t let this distract you. Your immediate goal is that the exponent on the $$\left( {z + 3} \right)$$ becomes 1. You will accomplish this by raising both sides of the equation to the reciprocal power of the $$\left( {z + 3} \right)$$. The reciprocal of 2/3 is 3/2. Therefore,
$$\left( {\left( {z + 3} \right)^{\large{\frac{2}{3}}} } \right)^{\large{\frac{3}{2}}} = \left( {5^{\large{\frac{1}{3}}} } \right)^{\large{\frac{3}{2}}} $$
Since we are raising an exponent to an exponent on each side, we multiply the exponents. This approach calls for caution, because by raising both sides to the 3/2 power, we are taking the square root (and cubing). To make sure that we account for all possible solutions, we must indicate ± on the square root.:
$$\left( {z + 3} \right) = \pm 5^{\large{\frac{3}{6}}} $$
So:
$$z + 3 = \pm 5^{\large{\frac{1}{2}}} $$
then,
$$z = \pm 5^{\large{\frac{1}{2}}} - 3$$
2. You are wise to think of these solutions as “potential solutions” and then to check to make sure they are solutions to the original equation. In this, a calculator can be helpful. Start with the original equation:
$$\left( {z + 3} \right)^{\large{\frac{2}{3}}} = 5^{\large{\frac{1}{3}}} $$ and then substitute each potential solution in turn.
$$( { - 5^{\large{\frac{1}{2}}} + 3} )^{\large{\frac{2}{3}}} \underbrace{\;=\;}_{?} 5^{{\large\frac{1}{3}}} $$ Both sides evaluate to 1.709975947.
$$
\left( {5^{\large{\frac{1}{2}}} + 3} \right)^{\large{\frac{2}{3}}} \underbrace{\;=\;}_{?} 5^{\large{\frac{1}{3}}}
$$ Both sides evaluate to 1.709975947.
Both possibilities are genuine solutions. Therefore, $$z = \pm 5^{\large{\frac{1}{2}}} - 3$$
Solve for x. That is, find the values of x that make the equation true.
$${\large\frac{2}{3}}\left( {x - 1} \right)^{ -{\large \frac{1}{3}}} = 0$$
1. Just as in Example 1, we have a number, $$\frac{2}{3}$$, multiplied by the entire quantity $$\left( {x - 1} \right)^{ -{\large \frac{1}{3}}} $$. So, we can multiply each side by the reciprocal of $$\frac{2}{3}$$ to get rid of it on the left hand side:
$${\Large\frac{3}{2}}\left( {{\Large\frac{2}{3}}\left( {x - 1} \right)^{ -{\large \frac{1}{3}} }} \right) = {\large\frac{3}{2}}\left( 0 \right)$$
This gives us:
$$\left( {x - 1} \right)^{ - {\large\frac{1}{3}}} = 0$$
2. Now, it is tempting to raise each side to the power of -3 to get rid of the exponent on the left hand side, but we cannot do that. Why not? A number raised to a negative exponent can be rewritten as a number raised to a positive exponent in the denominator of a fraction. So, what is the problem with raising each side to the exponent of -3? Well, on the right hand side, we would get:
$$0^{ - 3} = {\Large\frac{1}{{0^3 }}} = {\Large\frac{1}{0}}$$
And we CANNOT divide by zero!
Another way to think about this problem is by rewriting the left hand side with a positive exponent in the denominator. By doing this, we get:
$${\Large\frac{1}{{( {x - 1}}}} )^{\large{\frac{1}{3}}} = 0$$
Now, we have some unknown value in the denominator of a fraction which divides to give us 0. This is not possible. There is no value that we can divide out from 1 in order to get zero.
Thus, there are no solutions to this equation.
Some practice problems to check your skills:
Solve for the variable. That is, find the value(s) of the variable that make the equation true.
1. $$\left( {3x - 4} \right)^{\large{\frac{1}{4}} } = 2$$
Think about it for a moment and then access this link to view answer.
2. $$3\left( {z + 5} \right)^{\large{\frac{1}{5}}} = 2^{\large{\frac{1}{5}}} $$
Think about it for a moment and then access this link to view answer.
3. $$\left( {2 - x} \right)^{\large{\frac{2}{5}}} = 1$$
Think about it for a moment and then access this link to view answer.
4. $$\left( {2y + 5} \right)^{\large{\frac{4}{5}}} = 0$$
Think about it for a moment and then access this link to view answer.
5. $${\Large\frac{2}{7}}\left( {5x - 9} \right)^{\large{ - \frac{1}{7}}} = 0$$
Think about it for a moment and then access this link to view answer.