# Thread: How to solve this inequality?

1. ## How to solve this inequality?

Is there a way to solve for x in the equation: 5^(x-2) > -1?
It is 5 to the power of x-2.

I tried to take log on both sides but log a negative number doesn't seem right.

2. That is true for all x.

For all t, $\displaystyle 5^t>0$.

3. Ignore this - elementary mistake - see below

4. Originally Posted by e^(i*pi)
There are no real solutions because $\displaystyle 5^(x-2) \geq 0 \: , \: x \in \mathbb{R}$
In the complex plane the principle solution of $\displaystyle \ln(-1) \text{ is } i\pi$ (this follows from Euler's indentity - which happens to be my username)
$\displaystyle (x-2)\ln(5) > i\pi$
$\displaystyle x > \dfrac{2}{\ln(5)}+\dfrac{\pi}{\ln(5)}\,i$
What are you saying?
The inequality is true for every real number.

Moreover there is no ordering in the complex numbers.
So what does $\displaystyle x > \dfrac{2}{\ln(5)}+\dfrac{\pi}{\ln(5)}\,i$ mean?

5. Originally Posted by Plato
What are you saying?
The inequality is true for every real number.

Moreover there is no ordering in the complex numbers.
So what does $\displaystyle x > \dfrac{2}{\ln(5)}+\dfrac{\pi}{\ln(5)}\,i$ mean?
No idea how I missed that >.<

6. Originally Posted by Plato
That is true for all x.

For all t, $\displaystyle 5^t>0$.
ah thanks! how come I didn't think of that...haha...
thanks a lot!