Logarithms math problem?

(4^x-2)log(1-x^2)>0

=> (4^x)log(1-x^2) - 2log(1-x^2) > 0

=> (4^x)log(1-x^2) > 2log(1-x^2)

=> log((1-x^2)^(4^x)) > log((1 -x^2)^2)

=> (1-x^2)^(4^x) > (1 -x^2)^2

=> 4^x > 2

=> 2^(2x) > 2^1

=> 2x > 1

=> x > 1/2

However as the original equation includes log(1-x^2) and there are no real soltions of logs of values <= 0

so 1 - x^2 > 0

=> 1 > x^2

=> 1 > x > -1

as x > 1/2

=> 1 > x > 1/2

Think of logarithms as exponents. That's effectively what they are, and how they act. So treat the values as such.

You have an inequality, that, for simplicity's sake is

y*log(z) > 0 (using y and z to represent more complicated expressions).

This is saying, in effect, that log (z^y) > 0.

The value of any other number raised to the zero power is 1.

So the above expression really means;

z^y >1

Now treating this expression as an equality (so we can find the boundary case),

we have z^y = 1

Or (z^y)^-y = z = 1^-y (taking the yth root of each side).

But that's just saying that z = 1, since the yth root of 1 is still 1.

We could have saved ourselves some work - looking back at the original.

If we treat the > as an equality sign to discover the boundary condition, we notice that we could divide both sides by (4^x-2). Dividing 0 by anything still leaves zero. So the first factor doesn't play a role, at least not yet.

What we want is log(1-X^2) = 0 (again just mapping the inequality boundary)

That's the same as saying (1-x^2) = 1

Because again, a log is an exponent. So anything to the zero power =1

1-x^2 = 1

By inspection, x has to be zero to be true. that's the boundary. So when will the value of 1- x^2 be greater than 1? Never! (No real value of x will make 1-X^2 greater than 1. So the log will always be less that zero.

Negative log does NOT mean a negative number on the real number scale. In fact, at least not until complex variables, the log function is undefined except for positive real numbers. A negative log means least than 1 (fractional values).

This requires EITHER both 4^x - 2 >0 and log(1-x^2)>0

giving 2^(2x)>2 so 2x>1 so x>1/2 and 1-x^2>1 so x^2<0 which is

not possible.

OR 4^x - 2 <0 and log(1-x^2)<0

so x < 1/2 and 1-x^2 < 1 so x^2>0 so all x╪0

Complete set is x<1/2 except x=0

"It's LOG, it's LOG; it's big; it's heavy; it's wood..." See Ren and Stimpy if you don't understand...

By the way, I can't tell due to your syntax if it's 4^x - 2 or 4^(x - 2); I'll do both...

Remember that you cannot take the log of a negative number so in our example, -1 < x < 1 because otherwise (1 - x^2) < 0. Also, since x is within that boundary, log(1 - x^2) must be negative because log 1 = 0. For the left side to be positive, then, the first term must also be negative, so 4^x - 2 < 0 (under the first understanding), which means 4^x < 2 or 2^2x < 2^1, so 2x < 1 and x < 1/2. Properly written, it's -1 < x < 1/2.

If it's 4^(x - 2) then it has no solution because no value of x will make that negative, and it needs to be to make the inequality true. Therefore, I'm assuming that the former syntax is the correct one.