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please, solve 1/3log(x)-log10,000=4
\n" ); document.write( "..
\n" ); document.write( "1/3log(x)-log10,000=4
\n" ); document.write( "1/3log(x)-4=4
\n" ); document.write( "1/3log(x)=8
\n" ); document.write( "log (x^1/3)=8
\n" ); document.write( "x(1/3)=10^8 (exponential form)
\n" ); document.write( "cube both sides
\n" ); document.write( "x=10^24
\n" ); document.write( "..
\n" ); document.write( "the other question, is application of growth and decay.
\n" ); document.write( "here is the question, the half life of uranium,is 2.52(10^5)year. if 97.7% of the uranium in the original sample is present, what length of time is elapsed?(to the nearest thousand years.)
\n" ); document.write( "..
\n" ); document.write( "Formula for radioactive decay: A/Ao=2^-t/h, A=radioactive material present at time t, Ao=amt present initially, h=material half life.
\n" ); document.write( ".977=2^-t/2.52*10^5
\n" ); document.write( "take log of both sides
\n" ); document.write( "log.977=(-t/2.52*10^5)log2
\n" ); document.write( "(log.977)*(2.52*10^5)/log2=-t
\n" ); document.write( "-t≈-8460
\n" ); document.write( "length of time elapsed≈8000 years
\n" ); document.write( "..
\n" ); document.write( "solve this expression in terms of ln(x),
\n" ); document.write( "e^x=9
\n" ); document.write( "take log of both sides
\n" ); document.write( "xlne=ln9
\n" ); document.write( "lne=1
\n" ); document.write( "x=ln9
\n" ); document.write( "..
\n" ); document.write( "contract this expression
\n" ); document.write( "ln(3)-2(ln(4)+ln(8)
\n" ); document.write( "=ln[3*8/4^2]
\n" ); document.write( "=ln[24/16]
\n" ); document.write( "=ln[3/2]
\n" ); document.write( "=ln[1.5] \n" ); document.write( "