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Exponential growth and decay: what exactly does 0<b<1mean? does this mean that b is a decimal?
 2 years ago
 2 years ago
Exponential growth and decay: what exactly does 0<b<1mean? does this mean that b is a decimal?
 2 years ago
 2 years ago

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satellite73Best ResponseYou've already chosen the best response.0
no it means that b is between 0 and 1
 2 years ago

lilg132Best ResponseYou've already chosen the best response.0
which means it has to be a decimal
 2 years ago

lilg132Best ResponseYou've already chosen the best response.0
you cant have a whole number between 0 and 1
 2 years ago

satellite73Best ResponseYou've already chosen the best response.0
really? like for example 3.14?
 2 years ago

lilg132Best ResponseYou've already chosen the best response.0
how is 3.14 smaller then 1 and larger then 0?
 2 years ago

neverforgetvivisteeBest ResponseYou've already chosen the best response.0
and i don't really understand how the graph is supposed to be, how is it different from b > 1
 2 years ago

Geometry_HaterBest ResponseYou've already chosen the best response.1
no it has to be between 01 and 1
 2 years ago

satellite73Best ResponseYou've already chosen the best response.0
or perhaps the decimal \[\frac{1}{3}\]
 2 years ago

TBatesBest ResponseYou've already chosen the best response.0
I suppose it has to be of the form 1/a where a>1
 2 years ago

satellite73Best ResponseYou've already chosen the best response.0
point i was making is that 3.14 is a "decimal" but it is not less than 1 for that matter so is the decimal 100
 2 years ago

TBatesBest ResponseYou've already chosen the best response.0
But to get back to the actual question: if 0<b<1 then you have exponential decay: dw:1322013206618:dw If b>1 you have exponential growth: dw:1322013236762:dw
 2 years ago

neverforgetvivisteeBest ResponseYou've already chosen the best response.0
is it like 0. then whatever number?
 2 years ago

satellite73Best ResponseYou've already chosen the best response.0
0<b<1 means just what it says. that b is between 0 and 1. it may be represented by a decimal, or a fraction or maybe an irrational number like \[\ln(2)\]
 2 years ago

neverforgetvivisteeBest ResponseYou've already chosen the best response.0
oh so reading from left to right
 2 years ago

slaaibakBest ResponseYou've already chosen the best response.0
satellite, the 3.14 logic is flawed. the relevance of 3.14 being decimal has nothing to do with 0<b<1. but I understand your point on the fractions and logarithmic functions
 2 years ago

satellite73Best ResponseYou've already chosen the best response.0
"decimal " is not a synonym for numbers between 0 and 1. decimal just means base ten representation
 2 years ago

neverforgetvivisteeBest ResponseYou've already chosen the best response.0
so the decimal has to be 0.whatever number?
 2 years ago

neverforgetvivisteeBest ResponseYou've already chosen the best response.0
like 0.8 or 0.08302
 2 years ago

satellite73Best ResponseYou've already chosen the best response.0
no 5 is a decimal, 19.95 is a decimal
 2 years ago

neverforgetvivisteeBest ResponseYou've already chosen the best response.0
for the 0<b<1
 2 years ago

neverforgetvivisteeBest ResponseYou've already chosen the best response.0
are you there?
 2 years ago

slaaibakBest ResponseYou've already chosen the best response.0
I know. but the existence of a decimal larger than 1 does not influence the existence of a decimal less than 1 and greater than 0. just read your previous message about 3.14 closely. although this is a useless debate, the definition of a decimal was just misinterpreted and I agree with you on 0<b<1 not necessarily being decimal
 2 years ago

neverforgetvivisteeBest ResponseYou've already chosen the best response.0
but what kind of decimal does it have to be? something like 0.3 and 0.30823 or something like that?
 2 years ago

slaaibakBest ResponseYou've already chosen the best response.0
in base 10, yes. anything like 0.05878 or 0.99 or w/e. just less than 1 and more than 0
 2 years ago

neverforgetvivisteeBest ResponseYou've already chosen the best response.0
ok thanks to all of you who answered it's clear now
 2 years ago
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