$\log$

Algebra Level 3

If $\log{_b}{\sqrt{x}} = \sqrt{\log{_b}{x}}$ has two solutions, sum of which is 10, then $b^2+b-2$ is equal to

\sqrt3

\frac{1}{\sqrt{3}}

\sqrt{3}+1

\sqrt2

1 solution

Tommy Li
Aug 29, 2017

$\large \log{_b}{\sqrt{x}} = \sqrt{\log{_b}{x}}$

$\large \frac{1}{4}(\log_{b}x )^2-\log_{b}x = 0$

$\large \log_{b}x = 0 \ \text{or} \ \log_{b}x =4$

$\large x=1 \ \text{or} \ \large x=b^4$

$\large b^4+1 = 10$

$\large b = \sqrt[4]{9} = \sqrt{3}$

$\large b^2+b-2 = 3+\sqrt{3}-2 = \sqrt{3}+1$

Good solution

genis dude - 3 years, 9 months ago