# Thread: help solving an equation involving logs

1. ## help solving an equation involving logs

$\displaystyle \sqrt{log(x) - 3} = log(x) - 3$

I'm not sure how to solve this. Logically, it would seem to me that $\displaystyle log(x) - 3$ must equal 0 or 1 for $\displaystyle \sqrt{log(x) - 3}$ to equal $\displaystyle log(x) - 3$. I'm not sure how to prove it though. Could someone point me in the right direction?

2. Originally Posted by absvalue
$\displaystyle \sqrt{log(x) - 3} = log(x) - 3$

I'm not sure how to solve this. Logically, it would seem to me that $\displaystyle log(x) - 3$ must equal 0 or 1 for $\displaystyle \sqrt{log(x) - 3}$ to equal $\displaystyle log(x) - 3$. I'm not sure how to prove it though. Could someone point me in the right direction?
HI

You can let log x = b

$\displaystyle \sqrt{b-3}=b-3$ .. square both sides and with some algebra

$\displaystyle b^2-7b+12=0$

$\displaystyle (b-4)(b-3)=0$

b=4 or b=3

Then log x =4 or log x =3

Solve for x .

3. Thanks! That makes sense. So...

$\displaystyle \log(x) = 4 \Longleftrightarrow 10^4 = x$

or

$\displaystyle \log(x) = 3 \Longleftrightarrow 10^3 = x$

4. Originally Posted by absvalue
Thanks! That makes sense. So...

$\displaystyle \log(x) = 4 \Longleftrightarrow 10^4 = x$

or

$\displaystyle \log(x) = 3 \Longleftrightarrow 10^3 = x$
Absolutely

5. Originally Posted by absvalue
$\displaystyle \sqrt{log(x) - 3} = log(x) - 3$

I'm not sure how to solve this. Logically, it would seem to me that $\displaystyle log(x) - 3$ must equal 0 or 1 for $\displaystyle \sqrt{log(x) - 3}$ to equal $\displaystyle log(x) - 3$. I'm not sure how to prove it though. Could someone point me in the right direction?
Here is another attempt:

If $\displaystyle z = \sqrt{\log(x)-3}$

$\displaystyle z = z^2~\implies~z^2-z=0~\implies~z(z-1)=0~\implies~ z = 0~\vee~z=1$