document.write( "Question 1205369: if log10^2=0.3010 and log10 ^7=0.8451.Evaluate \"1\" log10 ^0.2 \"2\" log10^35^2 without using tables or calculator .Evaluate
\n" ); document.write( "non
\n" ); document.write( "

Algebra.Com's Answer #842115 by math_tutor2020(3817)\"\" \"About 
You can put this solution on YOUR website!

\n" ); document.write( "Problem 1\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "Each log shown below is base 10.
\n" ); document.write( "\"log%28%282%29%29+=+0.3010\" (given)
\n" ); document.write( "\"log%28%287%29%29+=+0.8451\" (given)\r
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "So,
\n" ); document.write( "\"log%28%280.2%29%29+=+log%28%282%2F10%29%29\"\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "\"log%28%280.2%29%29+=+log%28%282%29%29+-+log%28%2810%29%29\" Use log rule log(A/B) = log(A)-log(B)\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "\"log%28%280.2%29%29+=+0.3010+-+1\" Use the rule that log(10) = 1 where the log is base 10.\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "\"log%28%280.2%29%29+=+-0.699\" This is approximate.\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "-------------------------------------------------\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "Problem 2\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "Each log shown below is base 10.
\n" ); document.write( "\"log%28%2835%5E2%29%29+=+2%2Alog%28%2835%29%29\" Use log rule log(A^B) = B*log(A) to pull down the exponent.\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "\"log%28%2835%5E2%29%29+=+2%2Alog%28%285%2A7%29%29\" Factor 35 as 5*7.\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "\"log%28%2835%5E2%29%29+=+2%2A%28log%28%285%29%29%2Blog%28%287%29%29%29\" Use log rule log(AB) = log(A)+log(B)\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "\"log%28%2835%5E2%29%29+=+2%2A%28log%28%2810%2F2%29%29%2Blog%28%287%29%29%29\" Rewrite the 5 as 10/2.\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "\"log%28%2835%5E2%29%29+=+2%2A%28log%28%2810%29%29+-+log%28%282%29%29%2Blog%28%287%29%29%29\" Use log rule log(A/B) = log(A)-log(B)\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "\"log%28%2835%5E2%29%29+=+2%2A%281+-+0.3010%2B0.8451%29\" Plug in the given info; also log(10) = 1 when log is base 10.\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "\"log%28%2835%5E2%29%29+=+3.0882\" approximately\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "-------------------------------------------------\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "Another approach for problem 2\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "Each log shown below is base 10.
\n" ); document.write( "\"log%28%2835%5E2%29%29+=+2%2Alog%28%2835%29%29\" \r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "\"log%28%2835%5E2%29%29+=+2%2Alog%28%285%2A7%29%29\"\r
\n" ); document.write( "\n" ); document.write( "
\n" ); document.write( "\"log%28%2835%5E2%29%29+=+2%2Alog%28%28%2810%2F2%29%2A7%29%29\" \r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "\"log%28%2835%5E2%29%29+=+2%2Alog%28%287%2F%282%2F10%29%29%29\" \r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "\"log%28%2835%5E2%29%29+=+2%2A%28log%28%287%29%29+-+log%28%282%2F10%29%29%29\"\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "\"log%28%2835%5E2%29%29+=+2%2A%28log%28%287%29%29+-+log%28%280.2%29%29%29\"\r
\n" ); document.write( "\n" ); document.write( "
\n" ); document.write( "\"log%28%2835%5E2%29%29+=+2%2A%280.8451+-+%28-0.699%29%29\" Plug in log(7) = 0.8451, and use the result of problem 1.\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "\"log%28%2835%5E2%29%29+=+3.0882\" approximately
\n" ); document.write( "
\n" ); document.write( "
\n" );