How Do You Solve 10log1019 in Logarithms?

[m0286]

Hello sorry to post another so soon. This book is VERY hard to learn from... its not very good at explaining. The question says:
Evaluate the following logarithm: 10log1019. It never showed me how to do it when there's a number in front of the log. What I got from the log 1019 is log19/log10=1.27875. Now do I multiply this by 10 since that's the number infront.. I have no clue!
THNX AGAIN!

 

[whozum]

Yes, there's also a property of logarithms that applies here, there's probably a table of rules for the log function, this property will be there, it should be the only one with a coefficient infront of the function.

 

[chroot]

The ten out front has nothing to do with the log. The ten just multiplies the result of the log. You can, if you choose, apply a property of the logarithm which permits you to "absorb" the coefficient in front of the log into the log's argument, but there's absolutely no reason to do so in this case. (This property is what whozum is talking about.)

[tex]a \log_b x = \log_b x^a[/tex]

Once again, there's no reason to invoke this property unless you want to. I would suggest just multiplying the result of the logarithm by ten.

- Warren

 

[whozum]

I was merely pointing out that possibility. Perhaps he had seen it before and it would help him compute the log without the coefficient, although the answers are the same.

 

[dextercioby]

Incidentally,[itex]\log_{10} [/itex] also written [itex] \lg [/itex] is tabulated (or a specific function on a smart pocket calculator),so there's no need to change th basis.

Daniel.