Vanier College Practice test - Prove that $\log_{\frac {1} {\sqrt b} }\sqrt x= -\log_b x$

Solution 1:

You got stuck because you made a mistake: $\left(\frac {1} {\sqrt b} \right)^n \ne \frac {n} { (\sqrt b)^n}$

Your proof should go like this:

$n=\log_{\frac {1} {\sqrt b} }\sqrt x$

$\iff \left(\frac {1} {\sqrt b} \right)^n = \sqrt x $

$\iff \frac {1} { (\sqrt b)^n}= \sqrt x$

$\iff 1 = (\sqrt b)^n \sqrt x$

Can you finish your proof now?

Edit:

I see you finished your proof, but you did something more complicated. Your proof was almost done like this.

$\iff 1 = (\sqrt b)^n \sqrt x$

$\iff 1 = b^n x$

$\iff b^n = x^{-1}$

$\iff n = \log_b(x^{-1}) = -\log_b(x)$

Solution 2:

Hint: $\log_{\frac {1} {\sqrt b} }\sqrt x= -\log_\sqrt b \sqrt x=-\frac{\ln \sqrt x}{\ln \sqrt b}$