How do i solve: logx^2 = (logx)^2?

1. Oct 21, 2007

i do not know how to do this at ALL!!

2. Oct 21, 2007

Rogerio

2 * log (x) = log(x) ^ 2
Log(x) = 0 is root
OR
2 = log(x)
Then
x=1
OR
x=100

3. Oct 21, 2007

i dont get it....

4. Oct 21, 2007

quetzalcoatl9

$$\ln (x^2) = 2 \ln (x) = \ln (x) \ln (x)$$
$$2 = \ln (x)$$
$$x = e^2$$
and of course x=1

5. Oct 21, 2007

so is it like:

logx^2 = (logx)^2
2logx = (logx)(logx)
2 = logx
x = 10^2??? --> x = b^y = y = b^x ??? is that it??

6. Oct 21, 2007

do i subtract??
like:
2^y = -5^y-2 ???

7. Oct 22, 2007

Gib Z

You could think of it as a quadratic equation in log x.

8. Oct 22, 2007

HallsofIvy

Staff Emeritus
That's one difficulty with not showing any work! Is "log x" the common logarithm or the natural logarithm? In "elementary" work, it is standard to use "log" to mean the common logarithm (base 10) and "ln" to mean the natural logarithm (base e). In more advanced work, it is standard to use "log" to mean the natural logarithm and not use the common logarithm at all.

Rogerio was assuming your "log" was the common logarithm. Since your question made it look like you didn't know any thing about logarithms, he assumed it was "elementary" mathematics. Your answer, then used "log" to mean natural logarithms. The only way WE can know which is correct is for you to tell us and the only way for YOU to know is to check your text book.

9. Oct 22, 2007

Gib Z

Unless of course the answer is intended to be left in the log form =]

10. Jan 26, 2009

ronmathguy

this question is wrong
logx^2=2logx=logx+logx
and
(logx)^2= logx*logx
and
logx*logx =/= logx + log x

as simple as
x*y =/= x+y