Question 550001
{{{ log(16,4) = x }}}
You can rewrite this as
{{{ 16^x = 4 }}}
The square root of {{{16}}} is {{{4}}}, so
{{{ 16^(1/2) = 4 }}}
{{{ x = 1/2 }}}
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{{{ log(x,81) = 4 }}}
I can rewrite this as
{{{ x^4 = 81 }}}
{{{ 3^4 = 81 }}}
{{{ x = 3 }}}
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If you get confused, remember
" logs are exponents"
Whenever you have
log( base,a thing ) = another thing,
another thing is an exponent
with some base.