# Thread: More logs

1. ## More logs

1)log3(x^2-3x-1)=0

2)-logx(25)=-2
=x^-2=25

i know the base is 5 but confused whether it is negative or positive.

thanks

2. Originally Posted by smmmc
1)log3(x^2-3x-1)=0
$\displaystyle \log_b 1 = 0$ for b > 0, so
$\displaystyle \log_3 (x^2-3x-1) = 0$ becomes
$\displaystyle x^2-3x-1 = 1$.
\displaystyle \begin{aligned} x^2-3x-2 &= 0 \\ x &= \frac{3 \pm \sqrt{(-3)^2 - 4(1)(-2)}}{2(1)} \\ x &= \frac{3 \pm \sqrt{9 + 8}}{2} \\ x &= \frac{3 \pm \sqrt{17}}{2} \end{aligned}

2)-logx(25)=-2
=x^-2=25

i know the base is 5 but confused whether it is negative or positive.
$\displaystyle -\log_x 25 = -2$
First get rid of the negative signs:
$\displaystyle \log_x 25 = 2$.
As you said, x = 5 (and it's positive).

01

3. b>1 always apply ? what about when b<1 ?

thanks heaps

4. Oops, typo. I meant b > 0 and b ≠ 1. Of course, in logarithms you can't have a negative base (b < 0).

01

5. You can't have a base of 1 either

$\displaystyle b>0 \: , b \neq 1$

6. If the original problem had been
$\displaystyle log_x(25)= -2$
then it would have been $\displaystyle 25= x^{-2}= \frac{1}{x^2}$ so that $\displaystyle x^2= \frac{1}{25}$ and $\displaystyle x= \frac{1}{5}$, still positive.