SOLUTION: How do I solve logb(14b^2) given that logb(4)=1.39 and logb(7)=1.95
Thank you
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Question 1030367: How do I solve logb(14b^2) given that logb(4)=1.39 and logb(7)=1.95
Thank you
Found 2 solutions by josgarithmetic, solver91311:
Answer by josgarithmetic(39620) (Show Source): You can put this solution on YOUR website!
, and then you want to substitute the given information, but you have instead of .
Do some work on
which was originally given as....
NOW do the value substituteions.
and compute.
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
The log of the product is the sum of the logs, and \ =\ n\log_b(a))
, and \ =\ 1)
\ =\ \log_b(14)\ +\ \log_b(b^2))
\ +\ \log_b(7)\ +\ 2\log_b(b))
\ +\ \log_b(7)\ +\ 2)
But since
\ =\ \log_b(2^2)\ =\ 2\log_b(2))
and
\ =\ 1.39)
then it follows that
\ =\ \frac{1.39}{2})
Therefore:
\ =\ \frac{1.39}{2}\ +\ 1.95\ +\ 2)
You can do your own arithmetic.
John
My calculator said it, I believe it, that settles it
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