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) About Me  (Show Source):
You can put this solution on YOUR website!
log%28b%2C14%5E2%29

2%2Alog%28b%2C14%29

2%28log%28b%2C2%29%2Blog%28b%2C7%29%29, and then you want to substitute the given information, but you have log%28b%2C4%29 instead of log%28b%2C2%29.

Do some work on log%28b%2C4%29
log%28b%2C2%5E2%29
2%2Alog%28b%2C2%29
which was originally given as....
2%2Alog%28b%2C2%29=1.39
log%28b%2C2%29=%281%2F2%291.39
log%28b%2C2%29=0.695

NOW do the value substituteions.
2%280.695%2B1.95%29
and compute.

Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!


The log of the product is the sum of the logs, and , and







But since



and



then it follows that



Therefore:



You can do your own arithmetic.

John

My calculator said it, I believe it, that settles it