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A.Avinash_GouthamBest ResponseYou've already chosen the best response.0
loga  logb = log(a/b) loga + logb = log(a*b) x=log(e^x) logx=logy => x=y now help urself
 one year ago

LogicalAppleBest ResponseYou've already chosen the best response.1
dw:1356767087691:dw Can you think of the next step?
 one year ago

LogicalAppleBest ResponseYou've already chosen the best response.1
(That thing in the denominator of the parentheses is a 'b'  sorry for sloppy artwork)
 one year ago

BelleFlowerBest ResponseYou've already chosen the best response.0
I'm still not sure how to continue actually... :/ @LogicalApple
 one year ago

LogicalAppleBest ResponseYou've already chosen the best response.1
Consider e and ln as inverse functions. I.e., ln(e^x) = x, and e^(ln(x)) = x. Try raising e to the power of both sides. For example, on the left side you would get e^(ln K) = K
 one year ago

LogicalAppleBest ResponseYou've already chosen the best response.1
And whatever you do to one side you have to do to the other
 one year ago

BelleFlowerBest ResponseYou've already chosen the best response.0
What are inverse functions? Sorry i don't really get it
 one year ago

LogicalAppleBest ResponseYou've already chosen the best response.1
An inverse function cancels out the original function. Like square and square root functions are inverses of each other. Multiplication and division are inverses. Here, e and ln are inverses. That is why when we have e^(ln K), the e and ln cancel leaving only K.
 one year ago

BelleFlowerBest ResponseYou've already chosen the best response.0
But I am supposed to solve this just using the power, product, quotient laws and the changeofbase law....
 one year ago

LogicalAppleBest ResponseYou've already chosen the best response.1
But I see there is an 'e' in the answer, so you must be able to use the function.
 one year ago

LogicalAppleBest ResponseYou've already chosen the best response.1
I will present a solution. Tell me if you have any questions about it. One sec.
 one year ago
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