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

1) Evaluate: log_{6}6√6

I thought that by the power of a power law, I can make the logarithm:

= log_{6}(6)(6)^{0.5}

= 0.5 log_{6}(6)(6)

= 0.5 log_{6}(36)

= 0.5 log_{6}(6)^{2}

= (2)(0.5) log_{6}6

= 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 + log_{3}*x*^{2 }I really don't know what to do with this one, and the answer is log_{3}3x^{2}.

Please and thank you!

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

Quote:

Originally Posted by

**misiaizeska** 1) Evaluate: log_{6}6√6

$\displaystyle \log_6(6\sqrt{6})=\log_6(6)+0.5(\log_6(6))=1.5$

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:

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

2.) How can you rewrite 1 as a base 3 logarithm?

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

You could also have done this using the fact that $\displaystyle (6)(6^{0.5})= 6^{1+ 0.5}= 6^{1.5}$. And, of course, $\displaystyle log_6(6^x)= x$.