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
$\displaystyle \log_b 1 = 0$ for b > 0, so
$\displaystyle \log_3 (x^2-3x-1) = 0$ becomes
$\displaystyle x^2-3x-1 = 1$.
Solve the quadratic:
$\displaystyle \begin{aligned}
x^2-3x-2 &= 0 \\
x &= \frac{3 \pm \sqrt{(-3)^2 - 4(1)(-2)}}{2(1)} \\
x &= \frac{3 \pm \sqrt{9 + 8}}{2} \\
x &= \frac{3 \pm \sqrt{17}}{2}
\end{aligned}$
$\displaystyle -\log_x 25 = -2$2)-logx(25)=-2
=x^-2=25
i know the base is 5 but confused whether it is negative or positive.
First get rid of the negative signs:
$\displaystyle \log_x 25 = 2$.
As you said, x = 5 (and it's positive).
01