# Problems from Base change of the logarithms

Calculate the following logarithms:

$$log_3 15, \ log_5 \dfrac{1}{50}$$ and $$log_7 \sqrt[3]{147}$$

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### 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:

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

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