# How do I rewrite the expression $r+(3/2)s-(1/3)t$

Can someone please explain how to rewrite the expression $r+(3/2)s-(1/3)t$ in terms of $a, b,$ and $c$, where $r= \log (a), s=\log(b) \text{ and } t=2\log(c)$. The rewritten expression should be in the form of $\log (g(a,b,c))$

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Use $\log(x)+\log(y)=\log(xy)$, $\log(x)-\log(y)=\log(x/y)$, $k\log(x)=\log(x^k)$. –  lhf Aug 11 '11 at 11:46

I'm assuming that by "loge" you mean the natural logarithm with base $\mathrm e$, which I will write as "$\log$".
The key here are the relations $\log(xy)=\log x + \log y$ and $\log x^y=y\log x$. They allow you to rewrite as follows:
$$r+\frac32s-\frac13t=\log a+\frac32\log b-\frac132\log c=\log a+\log b^{3/2}+\log c^{-2/3}=\log(ab^{3/2}c^{-2/3})\;.$$