Question 1205369
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Problem 1


Each log shown below is base 10.
{{{log((2)) = 0.3010}}} (given)
{{{log((7)) = 0.8451}}} (given)



So,
{{{log((0.2)) = log((2/10))}}}


{{{log((0.2)) = log((2)) - log((10))}}} Use log rule log(A/B) = log(A)-log(B)


{{{log((0.2)) = 0.3010 - 1}}} Use the rule that log(10) = 1 where the log is base 10.


{{{log((0.2)) = -0.699}}} This is approximate.


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Problem 2


Each log shown below is base 10.
{{{log((35^2)) = 2*log((35))}}} Use log rule log(A^B) = B*log(A) to pull down the exponent.


{{{log((35^2)) = 2*log((5*7))}}} Factor 35 as 5*7.


{{{log((35^2)) = 2*(log((5))+log((7)))}}} Use log rule log(AB) = log(A)+log(B)


{{{log((35^2)) = 2*(log((10/2))+log((7)))}}} Rewrite the 5 as 10/2.


{{{log((35^2)) = 2*(log((10)) - log((2))+log((7)))}}} Use log rule log(A/B) = log(A)-log(B)


{{{log((35^2)) = 2*(1 - 0.3010+0.8451)}}} Plug in the given info; also log(10) = 1 when log is base 10.


{{{log((35^2)) = 3.0882}}} approximately


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Another approach for problem 2


Each log shown below is base 10.
{{{log((35^2)) = 2*log((35))}}} 


{{{log((35^2)) = 2*log((5*7))}}}

 
{{{log((35^2)) = 2*log(((10/2)*7))}}} 


{{{log((35^2)) = 2*log((7/(2/10)))}}} 


{{{log((35^2)) = 2*(log((7)) - log((2/10)))}}}


{{{log((35^2)) = 2*(log((7)) - log((0.2)))}}}

 
{{{log((35^2)) = 2*(0.8451 - (-0.699))}}} Plug in log(7) = 0.8451, and use the result of problem 1.


{{{log((35^2)) = 3.0882}}} approximately
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