Question

If logr 6 = m and logr 3 = n, then logr is equal to

Answer 1

If \[\log _{r} 6 = m, \log _{r} 3 = n\], then \[\log _{r} (r/2)\] is equal to?

Answer 2

First, notice that 6/3 = 2. So this means that if we subtract the given equations, we get


\[\large m-n=\log_{r}(6)-\log{r}(3)=\log_{r}\left(\frac{6}{3}\right)=\log_{r}(2)\]


So \[\large m-n=\log_{r}(2)\]

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Now let's simplify the given expression

\[\large \log_{r}\left(\frac{r}{2}\right)\]


\[\large \log_{r}\left(r\right)-\log_{r}\left(2\right)\]


\[\large 1-\log_{r}\left(2\right)\]


\[\large 1-(m-n)\]


\[\large 1-m+n\]

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This means that \[\large \log_{r}\left(\frac{r}{2}\right)=1-m+n\]

Answer 3

So for any logarithmic expression in which the base is the same as its factor the expression can be reduced to 1?

Answer 4

And thanks, by the way. The answer was correct

Answer 5

Exactly, for any number b (where b is not 1 and is nonnegative), the following is true


\[\large \log_{b}(b)=1\]

Answer 6

Ah, I don't know that. Thanks!