# Logarithm problem

• Oct 5th 2006, 10:55 AM
sarahh
Logarithm problem
Hi everyone,

Need a lot of help on this:

Find all real numbers x, if any, that satisfy the eq:

(log x^2)^2 = log x^4

NOTE: each log has base 10
• Oct 5th 2006, 11:15 AM
topsquark
Quote:

Originally Posted by sarahh
Hi everyone,

Need a lot of help on this:

Find all real numbers x, if any, that satisfy the eq:

(log x^2)^2 = log x^4

NOTE: each log has base 10

I'm assuming the LHS is [log(x^2)]^2 and the RHS is log(x^4)?

Then
[2 log(x)]^2 = 4 log(x)

Set y = log(x)

[2y]^2 = 4y

4y^2 = 4y

y^2 = y

y^2 - y = 0

y(y - 1) = 0

So y = 0 or y = 1.

Thus log(x) = 0 or log(x) = 1
log(x) = 0
10^[log(x)] = 10^0
x = 1

and
log(x) = 1
10^[log(x)] = 10^1
x = 10.

Thus x = 1 or x = 10.

-Dan
• Oct 5th 2006, 03:44 PM
Quick
Quote:

Originally Posted by topsquark
log(x) = 1
10^[log(x)] = 10^1
x = 10.

-Dan

What is the rule here...
• Oct 5th 2006, 03:53 PM
topsquark
Quote:

Originally Posted by Quick
What is the rule here...

The function
y = a^x
has an inverse defined as y = log(a) x. In English this reads the "log to the base a of x." (The "a" is subscripted.)

So we know that
a^[log(a) x] = x
and
[log(a) a^x] = x

Typically (though not universally, there's a thread on the forum somewhere on this) "log to the base 10" is usually shortened to the abreviation "log" (with no base mentioned) and "ln" is log(e) where e = 2.817...

-Dan

For general consumption, I fixed an error in my earlier post. (That's what I thought Quick was going to have posted about. :) )
• Oct 5th 2006, 04:41 PM
dan
good explanation topsquark. that even sence to me!!!

dan
• Oct 5th 2006, 06:09 PM
ThePerfectHacker
Quote:

Originally Posted by topsquark
The function
y = a^x

a>0 :eek:
• Oct 5th 2006, 11:55 PM
topsquark
Quote:

Originally Posted by ThePerfectHacker
a>0 :eek:

But of course!

-Dan
• Oct 5th 2006, 11:57 PM
earboth
Quote:

Originally Posted by ThePerfectHacker
a>0 :eek:

and a ≠ 1
• Oct 6th 2006, 12:08 AM
topsquark
Quote:

Originally Posted by earboth
and a ≠ 1

(laughs) You know I never thought about that one...but you're right.

-Dan
• Oct 6th 2006, 10:22 AM
ThePerfectHacker
Quote:

Originally Posted by earboth
and a ≠ 1

You can have exponents for 1 just it will be trivial.

You cannot have logarithms for 1, so that might be a reason.