Question 189805
{{{log(9,.3)}}}
This is saying that the base {{{9}}} raised to some
power gives me {{{.3}}}. I can rewrite that as:
{{{9^x = .3}}}
{{{9^x = 3/10}}}
{{{(3^2)^x = 3/10}}}
{{{3^(2x) = 3/10}}}
{{{10*3^(2x) = 3}}}
Now I can take the log to the base {{{10}}} of both sides
{{{1 + 2x*log(3) = log(3)}}}
{{{2x*log(3) = log(3) - 1}}}
{{{x = (log(3) - 1) / (2*log(3))}}}
{{{x = (.477121 - 1) / .954243}}}
{{{x = -.522879/.954243}}}
{{{x = -.547952}}}
check answer:
{{{9^-.547952 = .29999976}}}
This looks close enough
Actually,
{{{9^(-1/2) = 1/3}}} and not {{{3/10}}}