when I have .5=(0.91)^t, could i do
log(.5)/log(0.91) ?
3 answers
yes ... that would equal t
Your answer is correct, but you also can write:
log ( x ^ n ) = n ∙ log ( x )
0.5 = 1 / 2 = 1 / 2 ^ 1 = 2 ^ ( - 1 )
log ( 0.5 ) = log [ 2 ^ ( - 1 ) ] = ( - 1 ) * log ( 2 ) = - log ( 2 )
log ( 0.91 ^ t ) = t ∙ log ( 0.91 )
0.5 = 0.91 ^ t
log ( 0.5 ) = log ( 0.91 ^ t )
- log ( 2 ) = t ∙ log ( 0.91 ) Divide both sides by log ( 0.91 )
- log ( 2 ) / log ( 0.91 ) = t
t = - log ( 2 ) / log ( 0.91 )
log ( x ^ n ) = n ∙ log ( x )
0.5 = 1 / 2 = 1 / 2 ^ 1 = 2 ^ ( - 1 )
log ( 0.5 ) = log [ 2 ^ ( - 1 ) ] = ( - 1 ) * log ( 2 ) = - log ( 2 )
log ( 0.91 ^ t ) = t ∙ log ( 0.91 )
0.5 = 0.91 ^ t
log ( 0.5 ) = log ( 0.91 ^ t )
- log ( 2 ) = t ∙ log ( 0.91 ) Divide both sides by log ( 0.91 )
- log ( 2 ) / log ( 0.91 ) = t
t = - log ( 2 ) / log ( 0.91 )
ok thanks!