Question 241431
{{{log((root(3, x^2+21x)))=2/3}}}
As usual, there is more than one way to solve this. Perhaps the easiest solution is to recognize<ul><li>A cube root is the same as raising something to the 1/3 power.</li><li>There is a property of logarithms that allows you to move the exponent in the argument to the front as a coefficient: {{{log(a, (p^q)) = q*log(a, (p))}}}</li><li>We can multiply both sides of the equation by 3, once the 1/3 is a coefficient, and both the fractions will be eliminated. And eliminating fractions makes just about any equation eaier to solve.</li></ul>
Let's see this in action. Rewrite the cube root as am exponent of 1/3:
{{{log((x^2+21x)^(1/3))=2/3}}}
Use the previously mentioned property to move the exponent in front:
{{{(1/3)log((x^2+21x))=2/3}}}
Multiply both sides by 3:
{{{log((x^2+21x))=2}}}<br>
This is a much easier equation to solve than the one we had to start with. When the variable we are solving for is in the argument of a logarithm you often solve for the variable by rewriting the equation in exponential form. (When working with exponential and/or logarithmic equaitons it is important to remember how to switch from one form to the other using: {{{log(a, (p)) = q}}} is equivalent to {{{p = a^q}}}.) Rewriting our equation in logarithmic form we get:
{{{x^2 + 21x = 10^2}}}
which simplifies to:
{{{x^2 + 21x = 100}}}<br>
Now we have a quadratic equation to solve. So we want one side to be zero. Subtracting 100 from each side we get:
{{{x^2 + 21x - 100 = 0}}}
Now we factor (or use the Quadratic Formula):
{{{(x - 4)(x + 25) = 0}}}
According to the Zero Product Property this (or any) product can be zero only if one of the factors is zero. So:
{{{x - 4 = 0}}} or {{{x + 25 = 0}}}
Solving these we get:
{{{x = 4}}} or {{{x = -25}}}<br>
When solving logarithmic equations it is more than just a nice idea to check your answers. It is important. We have to make sure our possible answers do not make the argument of any logarithm zero or negative.<br>
{{{log((root(3, x^2+21x)))=2/3}}}
Checking x = 4:
{{{log((root(3, (4)^2+21(4))))=2/3}}}
{{{log((root(3, 16+21(4))))=2/3}}}
{{{log((root(3, 16+84)))=2/3}}}
{{{log((root(3, 100)))=2/3}}}
Since the cube root of a positive number is also positive, we know at this point that x = 4 does not make the argument of the log zero or negative. We can finish the check (to see if x=4 actually fits the equation) by rewriting in exponential form:
{{{root(3, 100) = 10^(2/3)}}}
Cubing both sides:
{{{(root(3, 100))^3 = (10^(2/3))^3}}}
{{{100 = 10^2}}}
{{{100 = 100}}} Check!<br>
Checking x = -25:
{{{log((root(3, (-25)^2+21(-25))))=2/3}}}
{{{log((root(3, 625+21(-25))))=2/3}}}
{{{log((root(3, 625 - 525)))=2/3}}}
{{{log((root(3, 100)))=2/3}}}
Since the cube root of a positive number is also positive, we know at this point that x = -25 does not make the argument of the log zero or negative. We can finish the check (to see if x = -25 actually fits the equation) by rewriting in exponential form:
{{{root(3, 100) = 10^(2/3)}}}
Cubing both sides:
{{{(root(3, 100))^3 = (10^(2/3))^3}}}
{{{100 = 10^2}}}
{{{100 = 100}}} Check!<br>In this case both solutions checked out.<br>
In other problems, if an x makes the argument of <b>any</b> logarithm either zero of negative, you must reject that solution. (Remember, it is the argument of log that cannot be zero or negative. It is OK for x to be zero or negative. For example, this problem had a negative solution, -25, that was correct.)<br>
Sometimes you may even reject all the answers you come up with. This would mean that your equation had no solutions at all.