# Thread: Exponential equation with modulus

1. ## Exponential equation with modulus

How to solve something like this?
$|3x-2|^{\sqrt{2x-3x^2}}=1$
$2x-3x^2=0$ doesn't work, because here x=2/3 and x=0, and the answer is 1/3...

2. The exponent will be 0 OR the base will be $\pm 1$.

$2x-3x^2 = 0$ which is $x=0$. x cannot be 2/3 because it would give the undefined $0^0$

You can also solve $|3x-2| = 1$ - one of the solutions is 1/3

edit: Not sure why 0 cannot be a solution because $f(0) = |3(0)-2|^{\sqrt{0-0}} = (2)^0 = 1$

3. e^(i pi) beat me to it. 8]
-Dan

4. The reason $x=2/3$ isn't working is that this gives you $0^0=1$, which I don't think is defined. Anyway, here is my very poor and untidy method, which does give the right answer in a 'round-the-houses' approach:

$|3x-2|^{\sqrt{2x-3x^2}}=1$
Split it into:

1) $(3x-2)^{\sqrt{2x-3x^2}}=1$
and
2) $(2-3x)^{\sqrt{2x-3x^2}}=1$

Take logs of both, starting with:

1) $(3x-2)^{\sqrt{2x-3x^2}}=1$

$\sqrt{2x-3x^2}\log(3x-2)=0$

Either $3x-2 = 1$ (this gives $log(1) = 0$)
In which case $x = 1$
Or as we've already established, $\sqrt{2x-3x^2}=0$
However neither work! The $x=1$ solution, when resubstituted, gives a complex value and is thus omitted. We've already gathered that the $\sqrt{2x-3x^2}=0$ doesn't work either. So then, we have to look at

2) $(2-3x)^{\sqrt{2x-3x^2}}=1$
Taking logs again:
$\sqrt{2x-3x^2}\log(2-3x)=Log(1)$
The only untested possibility is $2-3x=1$
In which case, $-1=-3x$
$x=\frac{1}{3}$

Interestingly, with my approach, the $x=0$ root indeed doesn't work - it gives negative values inside the logarithms, which are undefined, but that wasn't stated initially by the question. I assume whoever made the question took a similar approach as I did, or did something else, in which the 0 solution was lost along the way.

I know my method isn't sound, but it does explain the textbook's answer.

5. Quacky, I'm not sure what is the base for you logarithms? Do you mean $lg$?
I also don't understand why x can't be = 0. Because it is included in the domain of the equation. $x\in[0;\frac{2}{3}]$
And as e^(i*pi) wrote $f(0) = |3(0)-2|^{\sqrt{0-0}} = (2)^0 = 1$

6. Originally Posted by Evaldas
Quacky, I'm not sure what is the base for you logarithms? Do you mean $lg$?
Yes, that's what I mean, although the base of the log doesn't matter in this case, as long as it's consistent.

Originally Posted by Evaldas
And as e^(i*pi) wrote $f(0) = |3(0)-2|^{\sqrt{0-0}} = (2)^0 = 1$
I also don't understand why x can't be = 0. Because it is included in the domain of the equation. $x\in[0;\frac{2}{3}]$
I'm not sure either, as I said above, I supposed that someone followed a similar method as I did in which the 0 solution was accidentally dropped. Perhaps I didn't communicate my point very well.