SOLUTION: For some reason this question did not get posted the last time I asked, so I'm having to post it again.
In this problem I've been asked to prove that loga x = c * logb x, where
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-> SOLUTION: For some reason this question did not get posted the last time I asked, so I'm having to post it again.
In this problem I've been asked to prove that loga x = c * logb x, where
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Question 959994: For some reason this question did not get posted the last time I asked, so I'm having to post it again.
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.
I'm completely stumped here. Help! Answer by josgarithmetic(39620) (Show Source):
You can put this solution on YOUR website! This is not a statement to prove, unless you are allowed other important information: . This is a part of what you said, direct proportion or direct variation. The equation is true only because it is given as true.
Could you use Change of Base formula?
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If this is expected or given as true, then through their corresponding positions,
Now then you have a statement about c, but not much is proved here about the given equation.
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