# Logarithm square root problem

• Jan 3rd 2010, 05:46 AM
daparsons
Logarithm square root problem
Need some help with a log equation. I've done some searching and I found a very similar problem but didn't understand the response / solution.

$\sqrt{\log{x}}=\log{\sqrt{x}}$

This was the partial solution:
http://www.mathhelpforum.com/math-he...4b006d6c-1.gif

http://www.mathhelpforum.com/math-he...5be1e333-1.gif

http://www.mathhelpforum.com/math-he...4ff8d308-1.gif

set each factor equal to 0 and solve for x

I get the first two lines, but I don't see how we go from
http://www.mathhelpforum.com/math-he...5be1e333-1.gif to http://www.mathhelpforum.com/math-he...4ff8d308-1.gif

and I definitely don't get where to go from there. (Headbang)
Note, my equation uses base 10 logs, where the solution I found uses natural logs. That shouldn't matter, though, as I'm more interested in the mechanics of how to solve it than the actual solution.

Any help is greatly appreciated...

• Jan 3rd 2010, 06:02 AM
vonflex1
the transition from the first two lines onwards is

$\sqrt{lnx} * \sqrt {lnx} = lnx$

because:

$(lnx)^{0.5}*(lnx)^{0.5} = lnx^{0.5+0.5} = lnx$
• Jan 3rd 2010, 06:05 AM
Sudharaka
Dear daparsons,

Consider,

If you take squareroot(lnx) outside by using brackets you would get,

http://www.mathhelpforum.com/math-he...4ff8d308-1.gif

It is simple. However if you have any further queries please reply.
• Jan 3rd 2010, 06:10 AM
e^(i*pi)
Quote:

Originally Posted by daparsons
Need some help with a log equation. I've done some searching and I found a very similar problem but didn't understand the response / solution.

$\sqrt{\log{x}}=\log{\sqrt{x}}$

This was the partial solution:
http://www.mathhelpforum.com/math-he...4b006d6c-1.gif

http://www.mathhelpforum.com/math-he...5be1e333-1.gif

http://www.mathhelpforum.com/math-he...4ff8d308-1.gif

set each factor equal to 0 and solve for x

I get the first two lines, but I don't see how we go from
http://www.mathhelpforum.com/math-he...5be1e333-1.gif to http://www.mathhelpforum.com/math-he...4ff8d308-1.gif

and I definitely don't get where to go from there. (Headbang)
Note, my equation uses base 10 logs, where the solution I found uses natural logs. That shouldn't matter, though, as I'm more interested in the mechanics of how to solve it than the actual solution.

Any help is greatly appreciated...

Remember that $\sqrt{ln(x)} =0$ has no real (or complex to my knowledge) solutions
• Jan 3rd 2010, 06:57 AM
daparsons
Thanks, guys. I got it with a little substitution:

$\sqrt{\log{x}}=\log{\sqrt{x}}$
$\sqrt{u}=\frac{1}{2}u$
$2\sqrt{u}=u$
$4=u$
$\log{x}=4$
$x=10,000$

Quote:

Remember that http://www.mathhelpforum.com/math-he...c3f3eef0-1.gif has no real (or complex to my knowledge) solutions
Yeah, that's part of what was throwing me off. Now I understand how they got to the third line, but I still don't understand how to solve for x using that process. (Wondering)
• Jan 3rd 2010, 07:20 AM
Hello daparsons
Quote:

Originally Posted by daparsons
... I still don't understand how to solve for x using that process. (Wondering)

The principle is this:

• If the product of two (or more) expressions is equal to zero, then one (or more) of these expressions is itself equal to zero.

The most common example of this is the quadratic equation, where we factorise the quadratic expression (if possible), and then put each factor in turn equal to zero. For example:
$x^2-5x+6=0$

$\Rightarrow (x-2)(x-3)=0$
and it's when we put each of these factors in turn equal to zero:
$(x-2)=0$
or
$(x-3)=0$
that we get the solutions:
$x=2$
or
$x=3$
Well, we just apply the same principle here. When you get to:
$0 = \sqrt{\log x}(\tfrac12\sqrt{\log x} -1)$
you can then conclude that:
$\sqrt{\log x} = 0$
or
$(\tfrac12\sqrt{\log x} -1) = 0$
As has already been pointed out, the first of these possibilities, $\sqrt{\log x} = 0$, doesn't yield any values of $x$. So we're left with the conclusion that the only solution is when $\tfrac12\sqrt{\log x}-1 = 0$; i.e. $x = 10,000$.