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BelleFlower
 3 years ago
Please wait for image! Logarithms question
BelleFlower
 3 years ago
Please wait for image! Logarithms question

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

LogicalApple
 3 years ago
Best ResponseYou've already chosen the best response.1dw:1356767087691:dw Can you think of the next step?

LogicalApple
 3 years ago
Best ResponseYou've already chosen the best response.1(That thing in the denominator of the parentheses is a 'b'  sorry for sloppy artwork)

BelleFlower
 3 years ago
Best ResponseYou've already chosen the best response.0I'm still not sure how to continue actually... :/ @LogicalApple

LogicalApple
 3 years ago
Best ResponseYou've already chosen the best response.1Consider 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

LogicalApple
 3 years ago
Best ResponseYou've already chosen the best response.1And whatever you do to one side you have to do to the other

BelleFlower
 3 years ago
Best ResponseYou've already chosen the best response.0What are inverse functions? Sorry i don't really get it

LogicalApple
 3 years ago
Best ResponseYou've already chosen the best response.1An 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.

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

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

LogicalApple
 3 years ago
Best ResponseYou've already chosen the best response.1I will present a solution. Tell me if you have any questions about it. One sec.
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