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
$\log_b 1 = 0$ for b > 0, so
$\log_3 (x^2-3x-1) = 0$ becomes
$x^2-3x-1 = 1$.
\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.
$-\log_x 25 = -2$
First get rid of the negative signs:
$\log_x 25 = 2$.
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

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

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