How can i simplify this:
$\dfrac{\log_2 625}{\log_2 125}$
Thanks
How can i simplify this:
$\dfrac{\log_2 625}{\log_2 125}$
Thanks
Although you have now changed your question, the spirit of Thomas's answer is still correct. In particular, $\dfrac{log_2 625}{log_2 125} = \dfrac{log_5 625}{log_5 125} = \dfrac{4}{3}$. Does that make sense?
Without a change of base:
$\frac{\log_2 625}{\log_2 125} =\frac{\log_2 \left( 5^4\right)}{\log_2 \left(5^3\right)} =\frac{4 \log_2 5}{3 \log_2 5} =\frac{4 }{3 } .$
Here it is:
$\frac{\log_{2}625}{\log_{2}125} = \log_{125}625 = \frac{\log_{5}625}{\log_{5}125} = \frac{\log_{5}5^4}{\log_{5}5^3} = \frac{4}{3}$
by virtue of a basic property of logarithms. The basic property I am referring to is the following:
$ \frac{\log_{a}x}{\log_{a}y} = \log_{y}x $