document.write( "Question 10298: I want to solve the equation:
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document.write( "0.95 = exp(-a)+a*exp(-a) = exp(-a)*(1+a)\r
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document.write( "The first step I guess would be to take the natural log:
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document.write( "ln(0.95) = ln(1+a)-a\r
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document.write( "Please help!
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document.write( "Matt \n" );
document.write( "
Algebra.Com's Answer #5440 by mathmaven53(29)![]() ![]() ![]() You can put this solution on YOUR website! You can't solve for a algebaicly \n" ); document.write( "We try a numerical approach \n" ); document.write( " We have exp(-a)(1+a) = .95\r \n" ); document.write( "\n" ); document.write( " Then 1 + a = .95 exp(a)\r \n" ); document.write( "\n" ); document.write( " Plot the functions f(a) = 1+a and g(a) = .95 exp(a)\r \n" ); document.write( "\n" ); document.write( " See where they intersect and estimate an approximate value of a\r \n" ); document.write( "\n" ); document.write( " Call this value of a \r \n" ); document.write( "\n" ); document.write( " a_0\r \n" ); document.write( "\n" ); document.write( " Try iterating values in the recursion relation \r \n" ); document.write( "\n" ); document.write( " a_(n+1) = .95 exp(a_n) - 1\r \n" ); document.write( "\n" ); document.write( " Note: a_0 is read a sub zero, a_(n+1) is read a sub n plus one\r \n" ); document.write( "\n" ); document.write( " When you get values of this sequence that don't differ by much then quit \n" ); document.write( "the iterations. Then you have an approximate value of a \n" ); document.write( " |