Problems from Base change of the logarithms

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 .

$lo{g}_{3}15=lo{g}_3(3\cdot 5)=lo{g}_{3}3+lo{g}_{3}5=1+lo{g}_{3}5$

At this point, it is possible to apply the conversion. In this case, we will use the decimal logarithms, so:

$1+lo{g}_{3}5=1+{\displaystyle \frac{log5}{log3}}\simeq 1+{\displaystyle \frac{0,699}{0,477}}\simeq 1+1,465\simeq 2,465$

The same rule is valid for the second case, so that, on having decomposed $50$ into prime factors, we obtain $50=5^2\cdot 2$

The expression is simplified: $$lo{g}_5{\displaystyle \frac{1}{50}}=lo{g}_5{50}^{-1}=-1\cdot lo{g}_{5}50=-1\cdot lo{g}_5(5^2\cdot 2)=$$ $$=-2\cdot (lo{g}_{5}5+lo{g}_{5}2)=-2\cdot (1+{\displaystyle \frac{log2}{log5}})\simeq -2\cdot (1+{\displaystyle \frac{0,301}{0,699}})\simeq$$ $$\simeq -2\cdot (1+0,431)\simeq 2,862$$

Finally,

$lo{g}_7\sqrt[3]{147}$

Decomposing $147$ we obtain $147=7^2\cdot 3$

It is simplified: $$lo{g}_7\sqrt[3]{147}=lo{g}_7{147}^{\frac{1}{3}}={\displaystyle \frac{1}{3}}\cdot lo{g}_{7}147={\displaystyle \frac{1}{3}}\cdot lo{g}_7(7^2\cdot 3)=$$ $$={\displaystyle \frac{2}{3}}\cdot (lo{g}_{7}7+lo{g}_{7}3)={\displaystyle \frac{2}{3}}\cdot (1+{\displaystyle \frac{log3}{log7}})\simeq$$ $$=\simeq {\displaystyle \frac{2}{3}}\cdot (1+0,564)\simeq 1,043$$

Solution:

$2,465;2,862;1,043$

Hide solution and development