document.write( "Question 222790: Let (((log(a,X))))=C and (((log(b,X))))=D. Find the general statement that expresses (((log(ab,X))))in term of C and D.\r
\n" ); document.write( "\n" ); document.write( "I already know the answer is (CD)/(C+D)\r
\n" ); document.write( "\n" ); document.write( "I cannot figure out how to write a proof for this question. \r
\n" ); document.write( "\n" ); document.write( "I know all of the logarithmic principles as well but I can only come up with either (C)(D) as my answer or (C+D). I can't get a fraction.
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Algebra.Com's Answer #166740 by jim_thompson5910(35256)\"\" \"About 
You can put this solution on YOUR website!
I found that it helps to work backwards for a bit to find out how to go forward. In this case, I broke down \"log%28ab%2C%28x%29%29\" to get:\r
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\n" ); document.write( "\n" ); document.write( "\"log%28ab%2C%28x%29%29\" \r
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\n" ); document.write( "\n" ); document.write( "\"log%2810%2C%28x%29%29%2Flog%2810%2C%28ab%29%29\" (used the change of base formula here)\r
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\n" ); document.write( "\n" ); document.write( "\"log%2810%2C%28x%29%29%2F%28log%2810%2C%28a%29%29%2Blog%2810%2C%28b%29%29%29\" (used the identity \"log%28b%2C%28A%2AB%29%29=log%28b%2C%28A%29%29%2Blog%28b%2C%28B%29%29\")\r
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\n" ); document.write( "\n" ); document.write( "Then I realized that I needed to find substitutions for \"log%2810%2C%28a%29%29\" and \"log%2810%2C%28b%29%29\" (to somehow get 'x' in there). So I figured I'd solve for those respective expressions like so:\r
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\n" ); document.write( "\n" ); document.write( "\"log%28a%2C%28x%29%29=C\" Start with the first equation.\r
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\n" ); document.write( "\n" ); document.write( "\"log%2810%2C%28x%29%29%2Flog%2810%2C%28a%29%29=C\" Use the change of base formula.\r
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\n" ); document.write( "\n" ); document.write( "\"log%2810%2C%28x%29%29=C%2Alog%2810%2C%28a%29%29\" Multiply both sides by \"log%2810%2C%28a%29%29\".\r
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\n" ); document.write( "\n" ); document.write( "\"log%2810%2C%28x%29%29%2FC=log%2810%2C%28a%29%29\" Divide both sides by C.\r
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\n" ); document.write( "\n" ); document.write( "So this means \"log%2810%2C%28a%29%29=log%2810%2C%28x%29%29%2FC\"\r
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\n" ); document.write( "\n" ); document.write( "---------------------\r
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\n" ); document.write( "\n" ); document.write( "\"log%28b%2C%28x%29%29=D\" Start with the second equation.\r
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\n" ); document.write( "\n" ); document.write( "\"log%2810%2C%28x%29%29%2Flog%2810%2C%28b%29%29=D\" Use the change of base formula.\r
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\n" ); document.write( "\n" ); document.write( "\"log%2810%2C%28x%29%29=D%2Alog%2810%2C%28b%29%29\" Multiply both sides by \"log%2810%2C%28b%29%29\"\r
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\n" ); document.write( "\n" ); document.write( "\"log%2810%2C%28x%29%29%2FD=log%2810%2C%28b%29%29\" Divide both sides by D.\r
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\n" ); document.write( "\n" ); document.write( "So this means \"log%2810%2C%28b%29%29=log%2810%2C%28x%29%29%2FD\"\r
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\n" ); document.write( "\n" ); document.write( "-------------------------------------------\r
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\n" ); document.write( "\n" ); document.write( "Now let's get back to the main problem:\r
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\n" ); document.write( "\n" ); document.write( "\"log%28ab%2C%28x%29%29\" Start with the given expression.\r
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\n" ); document.write( "\n" ); document.write( "\"log%2810%2C%28x%29%29%2Flog%2810%2C%28ab%29%29\" Use the change of base formula.\r
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\n" ); document.write( "\n" ); document.write( "\"log%2810%2C%28x%29%29%2F%28log%2810%2C%28a%29%29%2Blog%2810%2C%28b%29%29%29\" Use the identity \"log%28b%2C%28A%2AB%29%29=log%28b%2C%28A%29%29%2Blog%28b%2C%28B%29%29\" to break up the denominator.\r
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\n" ); document.write( "\n" ); document.write( "Now here's where the substitutions come into play:\r
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\n" ); document.write( "\n" ); document.write( "\"log%2810%2C%28x%29%29%2F%28log%2810%2C%28x%29%29%2FC%2Blog%2810%2C%28x%29%29%2FD%29\" Plug in \"log%2810%2C%28a%29%29=log%2810%2C%28x%29%29%2FC\" and \"log%2810%2C%28b%29%29=log%2810%2C%28x%29%29%2FD\"\r
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\n" ); document.write( "\n" ); document.write( "From here, it's just boils down to algebraically simplifying the expression:\r
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\n" ); document.write( "\n" ); document.write( " Multiply the first inner fraction by \"D%2FD\".\r
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\n" ); document.write( "\n" ); document.write( " Multiply the second inner fraction by \"C%2FC\".\r
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\n" ); document.write( "\n" ); document.write( " Combine the lower fractions.\r
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\n" ); document.write( "\n" ); document.write( " Multiply \"log%2810%2C%28x%29%29\" by the reciprocal of the lower fraction.\r
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\n" ); document.write( "\n" ); document.write( " Multiply\r
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\n" ); document.write( "\n" ); document.write( "\"%28CD%2Alog%2810%2C%28x%29%29%29%2F%28log%2810%2C%28x%29%29%28D%2BC%29%29\" Factor out the GCF \"log%2810%2C%28x%29%29\" from the denominator.\r
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\n" ); document.write( "\n" ); document.write( " Cancel out the common terms.\r
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\n" ); document.write( "\n" ); document.write( "\"%28CD%29%2F%28D%2BC%29\" Simplify\r
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\n" ); document.write( "\n" ); document.write( "\"%28CD%29%2F%28C%2BD%29\" Rearrange the terms (trivial, but I figured that I'd match the book).\r
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\n" ); document.write( "\n" ); document.write( "So this means \"log%28ab%2C%28x%29%29=%28CD%29%2F%28C%2BD%29\" where \"log%28a%2C%28x%29%29=C\" and \"log%28b%2C%28x%29%29=D\"
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