SOLUTION: I dont know how to do this problem because of the square root, I couldnt even write it correctly on this post. Help! Given that {{{ log ( 2, 5 ) = 2.3219}}} and {{{ log ( 2, 7 ) =

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: I dont know how to do this problem because of the square root, I couldnt even write it correctly on this post. Help! Given that {{{ log ( 2, 5 ) = 2.3219}}} and {{{ log ( 2, 7 ) =      Log On


   



Question 108483: I dont know how to do this problem because of the square root, I couldnt even write it correctly on this post. Help!
Given that +log+%28+2%2C+5+%29+=+2.3219 and +log+%28+2%2C+7+%29+=+2.8074 , evaluate +log+%28+2%2C+square+root+35+%29+

Answer by Edwin McCravy(20059) About Me  (Show Source):
You can put this solution on YOUR website!
dont know how to do this problem because of the square root, I couldnt even write it correctly on this post. Help!
Given that +log+%28+2%2C+5+%29+=+2.3219 and +log+%28+2%2C+7+%29+=+2.8074 , evaluate +log+%28+2%2C+sqrt%2835%29%29+

First we change the square root to the 1%2F2 power,
since they are equivalent:

So the first step is to change 

+log+%28+2%2C+sqrt%2835%29%29+

to

+log+%28+2%2C+35%5E%281%2F2%29%29%29+

Second step is to move the exponent 1%2F2
in front of the logarithm as a coefficient:

1%2F2log+%28+2%2C+35%29%29+ 

Third step is to write 35 as 5%2A7 
So we now have

1%2F2log+%28+2%2C+5%2A7%29%29+

Fourth step, we put parentheses around the logarithm

1%2F2%28log+%28+2%2C+5%2A7%29%29%29+

Fifth step, we write the log%282%2C5%2A7%29 as log%282%2C5%29%2Blog%282%2C7%29
so now we have

1%2F2%28log%282%2C5%29%2Blog%282%2C7%29%29

Sixth step, we substitute the given decimal values for the logarithms,
that is, +log+%28+2%2C+5+%29+=+2.3219 and +log+%28+2%2C+7+%29+=+2.8074
so we now have

1%2F2%282.3219%2B2.8074%29

Seventh step, we evaluate the whole thing with a calculator:

Answer: 2.56465

Edwin