$\displaystyle \frac {2}{log_{5} 3}
= 2 \log_{3} 5
= 2 \frac{\log 5}{\log 3}$
what rules have been used in these steps?
In addition to the usual "rules of logarithms", you need the definition of logarithm.
If $\displaystyle x= log_3 5$, then $\displaystyle 3^x= 5$. Now take the logarithm, base 5, of that: $\displaystyle log_5 3^x= log_5 5$ and use the "law of logarithms" $\displaystyle log(a^b)= b log(a)$ to get [tex]xlog_5(3)= 1[tex] and so [tex]x= \frac{1}{log_5(3)}.
For the last take log of both sides of $\displaystyle 3^x= 5$: $\displaystyle log(3^x)= x log(3)= log(5)$ and divide both sides of the equation by log(3).
The "2" is just there to confuse things. Once you have those three different forms of "x" equal, multiply each by 2.
Hello, eq123!
. . . $\displaystyle \begin{array}{ccccccc}
[1] && [2] && [3] \\ \\[-3mm] \dfrac {2}{\log_5 3} & =& 2 \log_3 5 &=& 2\,\dfrac{\log 5}{\log 3} \end{array}$
What rules have been used in these steps?
Going from [1] to [2], the Reciprocal Rule is used: .$\displaystyle \log_b(a) \:=\:\frac{1}{\log_a(b)}$
Going from [2] to [3], the Base-Change Formula: .$\displaystyle \log_b(a) \:=\:\frac{\log(a)}{\log(b)} $