Simplify 7 log3 k + 6 log3 m − 9 log3 n. 4 log3 km over n 4 log3 (k + m − n) log3 k to the seventh power m to the sixth power over n to the ninth power log3 42 km over 9 n
These are fun! Do you still need help with this?
yes i do
Ok, since these all have the same base of 3 we can easily do this.
ok
When logs are added up there, that means that once upon a time, they were multiplied. When two logs are subtracted, that means that once upon a time, before they were expanded, they were divided. Let's start with the 7 the 6, and the 9 in front of all the terms.
how do we start
Those were exponents before they were expanded, so let's put them back that way. The rule for that is \[\log _{a}x ^{p}=p * \log _{a}x\]
Let's use that rule first:
\[\log _{3}(k)^{7}+\log _{3}(m)^{6}-\log _{3}(n)^{9}\]
See how that rule applied to what we had and now those numbers are written as exponents?
yes
k, now we will deal with the addition part. The rule for that is this:
\[\log _{a}x + \log _{a}y=\log _{a}(x * y)\]
That changes what we had as an added set of terms to:
\[\log _{3}(k ^{7} * m ^{6})-\log _{3}(n)^{9}\]
Can you follow that part using the multiplication property?
i cant
im bad at math
Ok, the log base is the same, so as long as the log base is the same, when you are adding two logs with the same bases, it's the same as multiplying them. That's just the way it is for logs; it doesn't have to make sense. That's just the rule.
Don't try to understand the WHY of these; they just ARE. "log base a of x + log base a of y= log base a of x * y." That's just the way it is.
would it be 4log3
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