document.write( "Question 631073: Letting log2=a, log3=b and log5=c, how do i express:
\n" ); document.write( "1. log2/3 in terms of a, b, and c
\n" ); document.write( "2. log5/3 in terms of a, b and c
\n" ); document.write( "3. log8/3........... in terms of A B and C? \n" ); document.write( "

Algebra.Com's Answer #397346 by jsmallt9(3758) $\"\"$ $\"About$

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You've been given the logs of 2, 3 and 5. You are also allowed to use well-known logs. With the base of \"log\" being 10, the log of any power of 10 (10, 100, 1000, 1/10, 1/100, etc.) are well-known and cn be used in these problems.

\n" ); document.write( "The keys to solving these problems is to rewrite the log in terms of the given and/or well-known logs. In this case we want to rewrite the logs in terms of log(2), log(3), log(5) and/or the log of any power of 10.

\n" ); document.write( "1. log(2/3)
\n" ); document.write( "We can see the 2 and the 3, There is a property of logs, $\"log%28a%2C+%28p%2Fq%29%29+=+log%28a%2C%28p%29%29+-+log%28a%2C+%28q%29%29\"$ , which allows us to apart the log of a quotient like this:
\n" ); document.write( "log(2/3) = log(2) - log(3)
\n" ); document.write( "Now that we have an expression in terms of log(2) and log(3), we can express this in terms a and b:
\n" ); document.write( "log(2/3) = log(2) - log(3) = a - b

\n" ); document.write( "2. log(5/3)
\n" ); document.write( "Again we can see the 5 and the 3, And it is a quotient again. So this problem is a lot like number 1:
\n" ); document.write( "log(5/3) = log(5) - log(3) = c - b

\n" ); document.write( "3. log(8/3)
\n" ); document.write( "We can see a 3. But a 2, 5 or power of 10 is not immediately obvious. To solve this problem we have to \"look inside\" the 8 trying to find 2's, 3's or powers of 10. With a little effort you should be able to see that $\"8+=+2%2A2%2A2+=+2%5E3\"$ So we start by expressing the 8 in terms of 2:
\n" ); document.write( " $\"log%28%288%2F3%29%29+=+log%28%282%5E3%2F3%29%29\"$
\n" ); document.write( "Now we can use the quotient property to split this in two:
\n" ); document.write( " $\"log%28%288%2F3%29%29+=+log%28%282%5E3%2F3%29%29+=+log%28%282%5E3%29%29+-+log%28%283%29%29\"$
\n" ); document.write( "For the first log we can use another property of logarithms, $\"log%28a%2C+%28p%5Eq%29%29+=+q%2Alog%28a%2C+%28p%29%29\"$ , to move the coefficient of the argument out in front:
\n" ); document.write( "

\n" ); document.write( "Now that we have an expression of log(2) and log(3) terms, we can substitute in the a and b:
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\n" ); document.write( "P.S. As it turned out, we never did need to use any of the well-known logs (i.e. logs of powers of 10). Just remember that in other problems we might need to use one or more of them. For example,
\n" ); document.write( "log(200/3) = log(200) - log(3) = log(100*2) - log(3) = log(100) + log(2) - log(3) = 2 + a - b \n" ); document.write( "

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