Express as a single logarithm.
log6496 – log62
I know logs pretty well but I don't know how to express 2 logs as a single log, can someone please help??

- anonymous

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- anonymous

I meant log base 6 of 496 - log base 6 of 2

- anonymous

\[\log_{6} 496-\log_{6} 2=\log_{6} \frac{ 496 }{ 2 }=\log_{6} 248\]

- anonymous

That's what I thought thank you! What about, log base 5 of 50 + log base 5 of 2.5?

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## More answers

- anonymous

I know it has to equal something in the final answer and that confuses me

- Nnesha

quotient rule\[\large\rm log_b x - \log_b y = \log_b \frac{ x }{ y}\]
to condense you can change subtraction to division
product rule \[\large\rm log_b x + \log_b y = \log_b( x \times y )\]
addition ----> multiplication
power rule \[\large\rm log_b x^y = y \log_b x\]
you should familiar with these

- Nnesha

there is a plus sign so which property you should apply ?

- anonymous

the product property

- Nnesha

yes right

- anonymous

I am doing a test review and all of the multiple choice answers for the question equal something, like one answer is, log base 5 of 125=3

- anonymous

How do you find what it equals?

- Nnesha

log base 5 of 50 + log base 5 of 2.5 for this question ?

- anonymous

yes

- Nnesha

\[\huge\rm log_5 50+ \log_5 2.5\] is this ur question ?

- anonymous

yes

- Nnesha

there isn't an equal sign..

- anonymous

These are the possible answers:
\[\log_{5}125=3 \]
\[\log_{5}112.5=2.93 \]
\[\log_{5}112.5=22.5 \]
\[\log_{5}125=625 \]

- Nnesha

alright then you have to apply the `change of base` formula \[\huge\rm log_b a= \frac{ \log a }{ \log b }\]

- Nnesha

\[\huge\rm log_5 50+ \log_5 2.5\] first) how would you write this in single log form b applying product property ?

- anonymous

\[\log_{5}125 \] ?

- Nnesha

yes right now use change of base formula

- anonymous

How do I know what to change the base to?

- Nnesha

base would stay the same hmm what do you mean ? o.O
here is an example \[\log_2 3 =\frac{ \log(3) }{ \log(2) }\]

- anonymous

oh okay! I was looking at my notes and I read the equation wrong!

- Nnesha

ohh

- anonymous

soooo is the answer \[\log_{5}125=3 \]

- Nnesha

if log_5 (125) is equal to 3 then yes
did you use the formula ?

- Nnesha

how would you use `change of base formula` for \[\large\rm \log_5 125\]

- anonymous

\[\frac{ \log(125) }{ \log(5) }\]

- Nnesha

perfect!

- anonymous

Awesome! This makes sense now! Thank you so much! You are very smart

- Nnesha

glad to hear that & thanks & good job!
good luck !

- anonymous

Thanks! :)

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