document.write( "Question 1191778: Solve for t in terms of q for the equation \"+q+=+1000%281%2F2%29%5E0.8%5Et+\" \n" ); document.write( "
Algebra.Com's Answer #823613 by Theo(13342)\"\" \"About 
You can put this solution on YOUR website!
i think i have it now.
\n" ); document.write( "i originally solved for q = 1000 * (1/2) ^ (.8 * t).
\n" ); document.write( "that was wrong.
\n" ); document.write( "the problem to be solved is q = 1000 * (1/2) ^ (.8 ^ t)
\n" ); document.write( "at least that's what i think it is now.
\n" ); document.write( "my calculator was not happy if there wasn't parentheses around the .8 ^ t.
\n" ); document.write( "so, .....
\n" ); document.write( "i am solving for q = 1000 * (1/2) ^ (.8 ^ t)\r
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\n" ); document.write( "\n" ); document.write( "start with q = 1000 * (1/2) ^ (.8 ^ t)
\n" ); document.write( "divide both sides of the equation by 10000 to get:
\n" ); document.write( "q / 1000 = (1/2) ^ (.8 ^ t)
\n" ); document.write( "take the log of both sides of the equation to get:
\n" ); document.write( "log(q/1000) = log((1/2)^(.8^t))
\n" ); document.write( "since log(a^b) = b * log(a), this becomes:
\n" ); document.write( "log(q/1000) = (.8^t)*log(1/2)
\n" ); document.write( "divide both sides by log(1/2) to get:
\n" ); document.write( "log(q/1000)/log(1/2) = .8^t
\n" ); document.write( "take the log of both sides of this equation to get:
\n" ); document.write( "log(log(q/1000)/log(1/2)) = log(.8^t)
\n" ); document.write( "since log(a^b) = b*log(a), this becomes:
\n" ); document.write( "log(log(q/1000)/log(1/2)) = t*log(.8)
\n" ); document.write( "divide both sides of the equation by log(.8) to get:
\n" ); document.write( "log(log(q/1000)/log(1/2))/log(.8) = t\r
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\n" ); document.write( "\n" ); document.write( "i confirmed by replacing t with 20 in q = 1000 * (1/2) ^ (.8 ^ t) and got:
\n" ); document.write( "q = 992.0404038.
\n" ); document.write( "i then replaced q with 992.0404038 in log(log(q/1000)/log(1/2))/log(.8) = t and got:
\n" ); document.write( "t = 20
\n" ); document.write( "i graphed both equations by replacing q with y and t with x to show that they are equivalent to each other.
\n" ); document.write( "the graph is shown below.\r
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\n" ); document.write( "\n" ); document.write( "the graph shows that the two equations are equivalent at x = 20
\n" ); document.write( "this means at t = 20 in the original equation.\r
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\n" ); document.write( "\n" ); document.write( "let me know if you have any questions.
\n" ); document.write( "theo
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