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

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?

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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