Calculate the following logarithms:
$$log_3 15, \ log_5 \dfrac{1}{50}$$ and $$log_7 \sqrt[3]{147}$$
Development:
We have to apply the rule of the logarithms conversion and other properties learned previously, but it is always good to firstly see if the expressions can be simplified a little bit.
Perhaps the numbers can be expressed using the base of the logarithm. To do this, it will be necessary to use, sometimes, decomposition in prime factors .
$$log_3 15=log_3 (3\cdot5)=log_3 3+log_3 5=1+log_3 5$$
At this point, it is possible to apply the conversion. In this case, we will use the decimal logarithms, so:
$$1+log_3 5=1+\dfrac{log5}{log3}\simeq 1+\dfrac{0,699}{0,477}\simeq 1+1,465\simeq 2,465$$
The same rule is valid for the second case, so that, on having decomposed $$50$$ into prime factors, we obtain $$50=5^2\cdot2$$
The expression is simplified: $$$log_5 \dfrac{1}{50}=log_5 50^{-1}=-1\cdot log_5 50=-1\cdot log_5 (5^2\cdot2)=$$$ $$$=-2\cdot(log_5 5+log_5 2)=-2\cdot\Big(1+\dfrac{log2}{log5}\Big)\simeq -2\cdot\Big(1+\dfrac{0,301}{0,699}\Big)\simeq$$$ $$$\simeq -2\cdot(1+0,431)\simeq2,862$$$
Finally,
$$log_7 \sqrt[3]{147}$$
Decomposing $$147$$ we obtain $$147=7^2\cdot3$$
It is simplified: $$$log_7 \sqrt[3]{147}=log_7 147^{\frac{1}{3}}=\dfrac{1}{3}\cdot log_7 147=\dfrac{1}{3}\cdot log_7 (7^2\cdot3)=$$$ $$$=\dfrac{2}{3}\cdot (log_7 7+log_7 3)=\dfrac{2}{3}\cdot\Big(1+\dfrac{log3}{log7}\Big)\simeq$$$ $$$=\simeq \dfrac{2}{3}\cdot(1+0,564)\simeq1,043$$$
Solution:
$$2,465; \ 2,862; \ 1,043$$