Question

log_{sqrt{3}} 9 + log_{25} 5

Answer 1

Hey Lisa :)
Hmm interesting base on that first one :d

Answer 2

\[\large\rm \log_{\sqrt3}9+\log_{25}5\]

Answer 3

Is that what the problem looks like?

Answer 4

This may help you http://www.purplemath.com/modules/logs.htm You can read that for a full description of what logs are and the relationship. What will be important for you is this relation: \[\huge { {y=b^x \rightarrow log_b(y)=x}}\]

Answer 5

yes that is what it looks like

Answer 6

sorry about the long response didn't realize how long I was away for

Answer 7

so, what is your b=? y=? for \(\log_{\sqrt3}9\)?

Answer 8

I'm not sure

Answer 9

@lisamath12 it turns out that \(\rm 9\) is the product of a bunch of \(\rm \sqrt3\)'s.
How many square root 3's would you multiply together to get 9?

\(\large\rm \sqrt3\sqrt3=3\)
Hmm so two of them is not enough.

Answer 10

\(\log_{\sqrt{3}} 9 + \log_{25} 5\)
Since \(\log_{\sqrt{3}} 9 = 2\log_{3} 3^2=2\times 2\log_33=4\)
Similarly \(\log_{25} 5 = 2\log_{5} 5=2\)
Hence \(\log_{\sqrt{3}} 9 + \log_{25} 5=4+2=6\)

Answer 11

Use the change of base formula\[\boxed{\log_ba = \frac{\log_c a}{\log_c b}}\]
\[\log_{\sqrt{3}} 9 + \log_{25} 5\\
= \frac{\log 9}{\log \sqrt 3}+\frac{\log5}{\log25}\\
= \frac{\log 3^2}{\log3^{1/2}}+\frac{\log5}{\log5^2}\]
Then use\[\boxed{\log_b a^n=n\log_b a}\]
\[= \frac{2\log 3}{\frac12\log3}+\frac{\log5}{2\log5}\]
cancel the logs
\[= \frac{2}{(\frac12)}+\frac12\]
simplify
\[=\]

Answer 12

\[\log_{\sqrt{3}} 9=x,\left( \sqrt{3} \right)^x=9,\left( 3 \right)^{\frac{ x }{ 2 }}=3^2,\frac{ x }{ 2 }=2,x=4\]
\[\log_{25} 5=y,25^y=5,\left( 5^2 \right)^y=5,5^{2y}=5^1,2y=1,y=\frac{ 1 }{ 2 }\]
add and get the result.