SOLUTION: log cube root(x^2+21x)=2/3 please be specific on how to solve. thank you

Question 241431: log cube root(x^2+21x)=2/3
please be specific on how to solve.
thank you
Answer by jsmallt9(3759) (Show Source): You can put this solution on YOUR website!

As usual, there is more than one way to solve this. Perhaps the easiest solution is to recognize

A cube root is the same as raising something to the 1/3 power.
There is a property of logarithms that allows you to move the exponent in the argument to the front as a coefficient:
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.

Let's see this in action. Rewrite the cube root as am exponent of 1/3:

Use the previously mentioned property to move the exponent in front:

Multiply both sides by 3:

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: is equivalent to .) Rewriting our equation in logarithmic form we get:

which simplifies to:

Now we have a quadratic equation to solve. So we want one side to be zero. Subtracting 100 from each side we get:

Now we factor (or use the Quadratic Formula):

According to the Zero Product Property this (or any) product can be zero only if one of the factors is zero. So:
or
Solving these we get:
or

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.

Checking x = 4:

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:

Cubing both sides:

Check!

Checking x = -25:

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:

Cubing both sides:

Check!
In this case both solutions checked out.

In other problems, if an x makes the argument of any 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.)

Sometimes you may even reject all the answers you come up with. This would mean that your equation had no solutions at all.
RELATED QUESTIONS
Please show how to do it. Solve the equation. log base 2 (2x+8) - log base 2... (answered by solver91311)

I need to solve this equation on a worksheet: log X +log (X + 2) = log 3 Thank... (answered by jim_thompson5910)

Please help me write this expression as a single logarithm:... (answered by josgarithmetic)

Simplify the expression.... (answered by lwsshak3)

please help me factor and solve x^2-21x-100=0 thank you...please show... (answered by user_dude2008)

Question: Express in terms of logarithms of x, y, and z: log [a] 4xy^5/z^5 Answers:... (answered by longjonsilver)

Solve the equation log 3√x^2+1 = 1/3. log (cube root... (answered by ikleyn)

please help me solve log base 6 X + log base 6 X-3=2 the answer is 7.365 but I need... (answered by lwsshak3)

please help me solve for x: {{{log(2,(log(3,x)))=4}}} thank... (answered by Alan3354)