document.write( "Question 959994: For some reason this question did not get posted the last time I asked, so I'm having to post it again.\r
\n" ); document.write( "\n" ); document.write( "In this problem I've been asked to prove that loga x = c * logb x, where c is a constant. In other words, direct proportionality. I'd started out by representing the left side of the above equation with p and the right with q, then went (left side)/p = 1 and (right side)/q = 1, therefore (left side)/p = (right side)/q. I then cross-multiplied and used the division property of equality to get... exactly what I already had, (left side)/p = (right side)/q. Just when I thought I had it cracked too! A later problem has provided a clue - if c were equal to loga b, that might be very helpful, but I can't figure out how to get there without making that a given.\r
\n" ); document.write( "\n" ); document.write( "I'm completely stumped here. Help!
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Algebra.Com's Answer #586697 by josgarithmetic(39620)\"\" \"About 
You can put this solution on YOUR website!
This is not a statement to prove, unless you are allowed other important information: \"log%28a%2Cx%29=c%2Alog%28b%2Cx%29\". This is a part of what you said, direct proportion or direct variation. The equation is true only because it is given as true.\r
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\n" ); document.write( "\n" ); document.write( "Could you use Change of Base formula?
\n" ); document.write( "\"log%28a%2Cx%29=log%28b%2Cx%29%2Flog%28b%2Ca%29\"
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\n" ); document.write( "\"log%28a%2Cx%29=log%28b%2Cx%29%2Flog%28b%2Ca%29=c%2Alog%28b%2Cx%29\"\r
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\n" ); document.write( "\n" ); document.write( "\"%281%2Flog%28b%2Ca%29%29%2Alog%28b%2Cx%29=c%2Alog%28b%2Cx%29\"
\n" ); document.write( "If this is expected or given as true, then through their corresponding positions,
\n" ); document.write( "\"highlight_green%281%2Flog%28b%2Ca%29=c%29\"
\n" ); document.write( "Now then you have a statement about c, but not much is proved here about the given equation.
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