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

[gibmeanswerz]

i do not know how to do this at ALL!

 

[Rogerio]

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

 

[gibmeanswerz]

i don't get it...

 

[quetzalcoatl9]

i do not know how to do this at ALL!

how about this:

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

 

[gibmeanswerz]

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??

 

[gibmeanswerz]

how about this: 2^y + 5^y-2??

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

 

[Gib Z]

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

 

[HallsofIvy]

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??

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 textbook.

 

[Gib Z]

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

 

[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