Base change of the logarithms

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: $$log_7 \sqrt[3]{147}=log_7 147^{\frac{1}{3}}=\dfrac{1}{3}\cdot log_7 147=\dfrac{1}{3}\cdot log_7 (7^2\cdot3)=$$ $$=\dfrac{2}{3}\cdot (log_7 7+log_7 3)=\dfrac{2}{3}\cdot\Big(1+\dfrac{log3}{log7}\Big)\simeq$$ $$=\simeq \dfrac{2}{3}\cdot(1+0,564)\simeq1,043$$