Logarithm Verification: Proving logbx = logax/logab for a, b > 0

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Homework Statement


Verify the formula logbx = logax/logab for a, b > 0


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The Attempt at a Solution


I don't even know how I would go about starting this...
 
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magnifik said:

The Attempt at a Solution


I don't even know how I would go about starting this...

If we let z=logbx, then what is 'x' equal to in terms of 'z' and 'b'?
 
rock.freak667 said:
If we let z=logbx, then what is 'x' equal to in terms of 'z' and 'b'?

x=bz
?
 
magnifik said:
x=bz
?

and if you take the loga of both sides of the equation, what does it reduce to?
 
omg thanks so much! lol that was a lot easier than i thought it would be
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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