Question 169332
if {{{y = x^n}}}
then {{{log(x,y) = n}}}
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let y = .125
let x = 2
let n = -3
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your equation becomes:
{{{log(2,(.125)) = -3}}}
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to prove this is correct, translate to base 10 and solve on calculator.
translation formula is:
{{{log(x,y)}}} = {{{(log(10,y))/(log(10,x))}}} = n
when x = 2, and y = .125, and n = -3, this formula becomes:
{{{log(2,(.125))}}} = {{{(log(10,(.125)))/(log(10,2))}}} = -3 = n
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solving on the calculator yields n = -3 which is correct.
basic equation for exponents and logarithms is:
{{{y = x^n}}} if and only if {{{log(x,y) = n}}}