Help me Simplify 9log9(7)
\[\log_{a}B^n=nlog_{a} B\]
9(log(9))(7) 60.117278 if this is wrong fell free to correct me
\(\log_ba = \dfrac{\ln a}{\ln b}\), I guess.
@aloud That's not simplifying though.
I i think it is 9^7
I don't think you should move 9 to exponent in log. Doesn't seem like "simplifying" to me.
Not sure, To me, your expression is already simplified. Unless you can do something about base.
after reworking it i get 7
Neither is transforming into a neiperian logarithm with a denominator which complicates the domain.
"complicates the domain" Do you even know what domain is?
Guys please i need help with this not argueing
You can hardly simplify a logarithm, you are very far from lecturing me.
Tell me, genius. How does simplifying logarithm change its domain?
@ganeshie8
Please help me......
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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: \[\log_{a}b^n=c <=>a^c=b^n\]
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 \(\log_9(7^9)\)
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
Do as you desire...
/thread
hahahaha... As I said, do as you deem fit, I have no desire to argue with unmature people.
Clearly because you don't want to make yourself look any more fool.
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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