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

##### 2 Answers

#### Answer:

#### Explanation:

#=> log_5 2 + log_5 25#

We know the value of

#=> 0.4307 + log25/log5#

Rewrite

#=> 0.4307 + log(5^2)/log(5^1)#

We now use the power rule that

#=> 0.4307 + (2log5)/(1log(5))#

#=> 0.4307 + 2#

#=> 2.4307#

Hopefully this helps!

#### Answer:

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

#### 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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Target is

Implying that

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Actual value of

Level of error

= say 1.04%

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