1. ## Logarithmic equation

For what value of is the following equation true?

$(log x )^3 = 4 log x$

So I need 3 X values I suppose and I need them rounded to 4 decimal places. So if I can figure out how to work this I should be fine. But I don't even know where to start the problem. Help would be greatly appreciated.

2. Originally Posted by n8thatsme
For what value of is the following equation true?

$(log x )^3 = 4 log x$

So I need 3 X values I suppose and I need them rounded to 4 decimal places. So if I can figure out how to work this I should be fine. But I don't even know where to start the problem. Help would be greatly appreciated.
lets see if this gives you any ideas. let $\log x = y$, you have

$y^3 = 4y$

can you solve for $y$?

3. I guess y would be equal to the cube root of 4y? I don't understand why I have 2 y's, should there just be one y and one x?

4. Originally Posted by n8thatsme
I guess y would be equal to the cube root of 4y? I don't understand why I have 2 y's, should there just be one y and one x?
that is not solving for y. to solve for y, you have to get y on one side of the equation by itself

$y^3 = 4y$

$\Rightarrow y^3 - 4y = 0$

$\Rightarrow y(y^2 - 4) = 0$

now what?

5. I got $y = 0, -2, 2$ but those aren't the correct answers.

6. Originally Posted by n8thatsme
I got $y = 0, -2, 2$ but those aren't the correct answers.
of course not, you were asked to solve for x, $y = \log x$, and thus it is not your solution, it is the log of your solutions

you must take the antilog of those values (if they exist)

7. We haven't learned antilogs yet. So I would plug in my y values into the equation so for example: y=0 it would be: $(0)^3=4(0)$?

8. Originally Posted by n8thatsme
We haven't learned antilogs yet. So I would plug in my y values into the equation so for example: y=0 it would be: $(0)^3=4(0)$?
no, what good would that do. we already know 0 = 0

i assure you that you have learned antilogs. these are just returning the logs to their exponential counterparts. logs are defined to make this easy transition

$\log_a b = c \Longleftrightarrow a^c = b$