Laws of Logarithms; how to evaluate logarithms/write as a single logarithms?

• November 17th 2012, 09:41 AM
misiaizeska
Laws of Logarithms; how to evaluate logarithms/write as a single logarithms?
1) Evaluate: log66√6
I thought that by the power of a power law, I can make the logarithm:
= log6(6)(6)0.5
= 0.5 log6(6)(6)
= 0.5 log6(36)
= 0.5 log6(6)2
= (2)(0.5) log66
= 1
However, the answer is 1.5 and I'm not sure where I went wrong; any corrections/hints?

2) Write each expression as a single logarithm; assume that all the variables represent positive numbers
1 + log3x2
I really don't know what to do with this one, and the answer is log33x2.

• November 17th 2012, 09:51 AM
Plato
Re: Laws of Logarithms; how to evaluate logarithms/write as a single logarithms?
Quote:

Originally Posted by misiaizeska
1) Evaluate: log66√6

$\log_6(6\sqrt{6})=\log_6(6)+0.5(\log_6(6))=1.5$
• November 17th 2012, 10:32 AM
MarkFL
Re: Laws of Logarithms; how to evaluate logarithms/write as a single logarithms?
1.) You may only bring the exponent out in front of the log function when the entire argument is raised to that exponent. To use this property in this problem, you could write:

$\log_6(6\sqrt{6})=\log_6(6^{\frac{3}{2}})=\frac{3} {2}\log_6(6)=\frac{3}{2}$

2.) How can you rewrite 1 as a base 3 logarithm?
• November 17th 2012, 10:51 AM
HallsofIvy
Re: Laws of Logarithms; how to evaluate logarithms/write as a single logarithms?
You could also have done this using the fact that $(6)(6^{0.5})= 6^{1+ 0.5}= 6^{1.5}$. And, of course, $log_6(6^x)= x$.