SOLUTION: Solve for t in terms of q for the equation {{{ q = 1000(1/2)^0.8^t }}}

Question 1191778: Solve for t in terms of q for the equation
Found 3 solutions by Theo, josgarithmetic, ikleyn:
Answer by Theo(13342)   (Show Source): You can put this solution on YOUR website!
i think i have it now.
i originally solved for q = 1000 * (1/2) ^ (.8 * t).
that was wrong.
the problem to be solved is q = 1000 * (1/2) ^ (.8 ^ t)
at least that's what i think it is now.
my calculator was not happy if there wasn't parentheses around the .8 ^ t.
so, .....
i am solving for q = 1000 * (1/2) ^ (.8 ^ t)

start with q = 1000 * (1/2) ^ (.8 ^ t)
divide both sides of the equation by 10000 to get:
q / 1000 = (1/2) ^ (.8 ^ t)
take the log of both sides of the equation to get:
log(q/1000) = log((1/2)^(.8^t))
since log(a^b) = b * log(a), this becomes:
log(q/1000) = (.8^t)*log(1/2)
divide both sides by log(1/2) to get:
log(q/1000)/log(1/2) = .8^t
take the log of both sides of this equation to get:
log(log(q/1000)/log(1/2)) = log(.8^t)
since log(a^b) = b*log(a), this becomes:
log(log(q/1000)/log(1/2)) = t*log(.8)
divide both sides of the equation by log(.8) to get:
log(log(q/1000)/log(1/2))/log(.8) = t

i confirmed by replacing t with 20 in q = 1000 * (1/2) ^ (.8 ^ t) and got:
q = 992.0404038.
i then replaced q with 992.0404038 in log(log(q/1000)/log(1/2))/log(.8) = t and got:
t = 20
i graphed both equations by replacing q with y and t with x to show that they are equivalent to each other.
the graph is shown below.

the graph shows that the two equations are equivalent at x = 20
this means at t = 20 in the original equation.

let me know if you have any questions.
theo
Answer by josgarithmetic(39620)   (Show Source): You can put this solution on YOUR website!

Another solution method could be better.
Answer by ikleyn(52798)   (Show Source): You can put this solution on YOUR website!
.

I am 129% sure that the formula in your post is written/presented INCORRECTLY.

Double check it.

If necessary, re-post to the forum after revising.

Please to not post it to me personally.

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