Question 202363
your original equation is:
{{{(125/27)^x = (.6)^(1/3)}}}
by the laws of logarithms, this can only be true if and only if
{{{log((125/27),((.6)^(1/3))) = x }}}
we can convert this log with a base of (125/27) to a log with a base of 10 so that we can solve it using the calculator.
the logarithm base conversion formula is:
{{{log((125/27),((.6)^(1/3))) = log(10,((.6)^(1/3))) / log(10,(125/27))}}}
our formula now becomes:
{{{ log(10,((.6)^(1/3))) / log(10,(125/27)) }}} = x.
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now:
{{{log(10,((.6)^(1/3)))}}} is the same as {{{(1/3)*(log(10,(.6)))}}}.
also:
{{{ log(10,(125/27)) }}} is the same as {{{ log(10,125) - log(10,27) }}}
so the equation becomes:
x = {{{((1/3)*(log(10,(.6))))}}} / {{{ (log(10,125) - log(10,27)) }}}
you can now use your calculator to solve this.
the equation becomes:
x = {{{((1/3)*(-.22184875))/((2.096910013)-(1.431363764))}}}
which becomes:
x = {{{(-0.073949583)/(.665546249)}}}
which becomes:
x = -0.1111111111111
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to prove your answer is correct, substitute it in the original equation to get:
{{{(125/27)^(-0.111111111111) = (.6)^(1/3)}}}
this becomes:
{{{0.843432665 = .843432665}}}
which is true.
note that .843432665 = (.6)^(1/3)
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