SOLUTION: 2log3+log2x-log(x+1)
I can't figure out how to write this as a single logarithm.
Thank you!
Algebra.Com
Question 382269: 2log3+log2x-log(x+1)
I can't figure out how to write this as a single logarithm.
Thank you!
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
\ +\ \log(2x)\ -\ \log(x\ +\ 1))
Use
\ =\ n\log_b(x))
to write:
\ +\ \log(2x)\ -\ \log(x\ +\ 1))
\ +\ \log(2x)\ -\ \log(x\ +\ 1))
Use "The sum of the logs is the log of the product"
 + \log_b(y) = \log_b(xy))
to write:
\ -\ \log(x\ +\ 1))
Use: "The difference of the logs is the log of the quotient"
 - \log_b(y) = \log_b\left(\frac{x}{y}\right))
to write:
)
And that, as they say, is that.
John
My calculator said it, I believe it, that settles it
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