Help me Simplify 9log9(7)

- anonymous

Help me Simplify 9log9(7)

- schrodinger

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- Owlcoffee

\[\log_{a}B^n=nlog_{a} B\]

- anonymous

9(log(9))(7)
60.117278
if this is wrong fell free to correct me

- geerky42

\(\log_ba = \dfrac{\ln a}{\ln b}\), I guess.

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## More answers

- geerky42

@aloud That's not simplifying though.

- anonymous

I i think it is 9^7

- geerky42

I don't think you should move 9 to exponent in log. Doesn't seem like "simplifying" to me.

- geerky42

Not sure, To me, your expression is already simplified. Unless you can do something about base.

- anonymous

after reworking it i get 7

- Owlcoffee

Neither is transforming into a neiperian logarithm with a denominator which complicates the domain.

- geerky42

"complicates the domain"
Do you even know what domain is?

- anonymous

Guys please i need help with this not argueing

- Owlcoffee

You can hardly simplify a logarithm, you are very far from lecturing me.

- geerky42

Tell me, genius. How does simplifying logarithm change its domain?

- geerky42

- anonymous

Please help me......

- geerky42

|dw:1434463121313:dw|

- Owlcoffee

In spite of geerky, who clearly hasn't mastered basic algebra yet.
We will use the property I stated above:
\[\log_{a} B^n=nlog_{a} B\]
So, for the problem:
\[9\log_{9} (7)\]
We'll just take it to the exponent:
\[\log_{9}(7)^9 \]
That would be a more simplified version, because later on, you'll learn that log(a)B^n is a more simplified way of expressing a logarithm because of the very definition:
\[\log_{a}b^n=c <=>a^c=b^n\]

- geerky42

We are simplifying a expression, not equation. So turn it into exponential form wouldn't work out well.
Plus it's not \(\log_9(7)^9\), it's \(\log_9(7^9)\)

- geerky42

Multiplication looks simplified to me than exponent.
So as far as I concern, given expression is pretty much already simplified.
Exactly what kind of form are you asked to simplify into? @maryah.g.osborn

- Owlcoffee

Do as you desire...

##### 1 Attachment

- geerky42

http://www.wolframalpha.com/input/?i=9+log+base+9+%287%29

- geerky42

/thread

- Owlcoffee

hahahaha... As I said, do as you deem fit, I have no desire to argue with unmature people.

- geerky42

Clearly because you don't want to make yourself look any more fool.

- ganeshie8

relax friends
i think the question looks terrible because of not using latex, i think it should be
\[\large 9^{\log_9 7}\]
@maryah.g.osborn please confirm if the expression looks as above or attach a screenshot if possible

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