Question 364793
This can be understood has working a rule backwards:
rule:
{{{log(a,b^c) = c*log(a,b)}}}
So, what you have is:
{{{a = 6}}}
{{{b = 5x}}}
{{{c = -2}}}
{{{-2*log(6,5x) = log(6,(5x)^(-2))}}}
{{{ log(6,(5x)^(-2)) = log(6,(1/(25x^2)))}}}