I'm stuck on solving (Log X)^2 + Log X ^5 = -6

[traunit]

Solve For X:

(Log X) ^2 + Log X ^5 = -6

2nd Step (Log X) (Log X) + Log X^5 = -6 is this correct?

Next step....

 

[pka]

\(\displaystyle \L\begin{array}{l}
\left( {\log (x)} \right)^2 + \log \left( {x^5 } \right) = - 6\quad \Rightarrow \quad \left( {\log (x)} \right)^2 + 5\log \left( x \right) + 6 = 0 \\
Z = \log (x)\quad \Rightarrow \quad Z^2 + 5Z + 6 = 0 \\
\end{array}.\)

Solve for Z. Then solve for x.

 

[traunit]

Why did you put Z=Log X?

Can't you just solve it without doing that?

 

[Mrspi]

Why did you put Z=Log X?

Can't you just solve it without doing that?

Well, sure you could.....but temporarily replacing log x with z makes the factoring easier for many people.

 

[traunit]

After I factored I got:

x= -2 and x=-3

I put Log -2 and it copmes up error in my calculator so does -3.

Did I do this right?

 

[pka]

After I factored I got: x= 2 and x=3

No you did not!
You got Z=+2 or log(x)=2.
So what is x?

 

[traunit]

I don't understand how you got x=2

 

[Mrspi]

I don't understand how you got x=2

We had
z^2 + 5z + 6 = 0

Factor the left side:
(z + 3)(z + 2) = 0

Set each factor equal to 0 and solve:
if z + 3 = 0, then z = -3
if z + 2 = 0, then z = -2

Now, remember that

z = log x

So.....if z = -3, we have

-3 = log x
Change to exponential form:
10<SUP>-3</SUP> = x

0.001 = x

Now, you do the same thing for z = -2.......

 

[traunit]

so 10^-2= .01

so x= .001 and x=.01?

Would the problem completely change if it was written

(Log )^2 =Log ^5=-6
x x