SOLUTION: Given log2on the base 10 =.3010, find log200on the base25 by using log table

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Question 970606: Given log2on the base 10 =.3010, find log200on the base25 by using log table
Answer by Theo(13342)   (Show Source): You can put this solution on YOUR website!
not exactly sure what you mean by this, but one way to solve this is as follows:

in the following:

big L means log and L(x) means log of x to the base of 10 and L25(x) means log of x to the base of 25.

using that convention:

log(2) on the base of 10 is represented as L(2).

log(200) on the base of 25 is represented as L25(200) which can be converted to base of 10 to make it equal to L(200)/L25).

so you get L(2) = .3010 and you want to find the value of L(200)/L(25)

L(200)/L(25) can be expressed as L(2*100)/L(100/4).

this can be expressed as (L(2) + L(100)) / (L(100 - L(4)).

this can be expressed as (L(2) + L(10^2)) / (L10^2) - L(2^2)).

this can be expressed as (L(2) + 2*L(10)) / (2*L(10) - 2*L(2))

you are given that L(2) = .3010.
actually L(2) is equal to .3010299957.

using the unrounded number makes confirming your solution easier down the road, so i'll use that instead of .3010.

since L(10) is equal to y if and only if 10^y = 10 if and only y = 1, then you get:

L(10) = 1.

your expression of (L(2) + 2*L(10)) / (2*L(10) - 2*L(2)) becomes:

(.3010299957 + 2*1) / (2*1 - 2*.3010299957) which becomes:

(.3010299957 + 2) / (2 - 2*.3010299957) which becomes:

2.3010299957 / 1.397940009 which becomes 1.646014837

you can confirm this is true by using your calculator to get L(200)/L(25) = 1.646014837

since they're the same, the solution is good.




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