How do you approximate log_5 50 given log_5 2=0.4307 and log_5 3=0.6826?

Dec 21, 2016

${\log}_{5} 50 \cong 2.4307$

Explanation:

${\log}_{5} 50$ can be written as ${\log}_{5} \left(2 \times 25\right)$, which can be rewritten as ${\log}_{a} \left(n \times m\right) = {\log}_{a} n + {\log}_{a} m$.

$\implies {\log}_{5} 2 + {\log}_{5} 25$

We know the value of ${\log}_{5} 2$ is $0.4307$ but to find the value of ${\log}_{5} 25$ we must use the change of base rule such that ${\log}_{a} n = \log \frac{n}{\log} a$.

$\implies 0.4307 + \log \frac{25}{\log} 5$

Rewrite $25$ and $5$ in powers of $5$.

$\implies 0.4307 + \log \frac{{5}^{2}}{\log} \left({5}^{1}\right)$

We now use the power rule that $\log {a}^{n} = n \log a$.

$\implies 0.4307 + \frac{2 \log 5}{1 \log \left(5\right)}$

$\implies 0.4307 + 2$

$\implies 2.4307$

Hopefully this helps!

Dec 21, 2016

The question is what I call 'a little bit wooly'!

${\log}_{5} 50 \approx 2.4054$ with 1.04% error

Explanation:

The inclusion of the word 'approximate' implies a rather wider range of precision than is normally generated by standard calculations. There is a problem hear in as much as 'approximate' is rather open to interpretation. I bit like "how long is a piece of string"!
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color(red)("Assumption: we are meant to use both of "log_5 3" and "log_5 2

Target is $\text{ } {\log}_{5} \left(50\right)$

$50 \to 3 \times {2}^{4} = 48$ as an approximation of 50

Implying that

${\log}_{5} 50 \text{ is somewhere near to } {\log}_{5} 3 + {\log}_{5} {2}^{4}$

$\implies {\log}_{5} 3 + 4 {\log}_{5} 2$

$\implies 0.6826 + \left(4 \times 0.4307\right) \text{ "=" } 2.4054$

${\log}_{5} 50 \approx 2.4054$
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Actual value of $\text{ "log_5 50 =(log_10 50)/(log_10 5) = 2.4307" to 4dp}$

Level of error$\to 2.4307 - 2.4054 = 0.0253 \pm 0.00005$

= say 1.04%
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${\log}_{5} 50 \approx 2.4054$ with 1.04% error