Hello there, i received my exam back i got this question wrong well there is two of them maybe someone could show me how to do it?
1)logE=log(10.61)+0.1964logm ----------- E= Eye axial length m= body mass in kilograms if E = 31 solve for m round to closest kilogram i got 8.57 and it was wrong.
2)log36^5x=log216^x-7------------------solve for x ( what i tried doing was bringing the exponents down through the power law and solve using quotient rules and it was wrong
any help would be appreciated.
so you want to solve 0.1964 log(m)+ log(10.61)= log(31).
Then 0.1964 log(m)= log(31)- log(10.61)= log(31/10.61)= log(2.922) so log(m)= log(2.922)/0.1964= log(2.922^(1/0.1964))= log(2.922^5.092) and then m= 2.922^5.092= 235.1, nowhere near "8.57".
Assuming you mean log(35^(5x))= log(216^(x- 7)), yes, that is 5x log(35)= (x- 7)log(216)= log(216)x- 7log(216), x(5 log(35)- log(216))= -7log(216). x= (-7 log(216))/(5log(35)- log(216)).2)log36^5x=log216^x-7------------------solve for x ( what i tried doing was bringing the exponents down through the power law and solve using quotient rules and it was wrong
any help would be appreciated.
So does anyone know how to proceed from here log(m)=log(2.922)/.1964 after this step i am lost, i don't know how it turns into 2.922^1/0.1964 for pt 2?
And thanks Plato for the coding tip, i still dont know how to use LaTeX never used any code in my life.
This subforum will help you with the code. Once you begin, you quickly learn the code.
If you click on the “go advanced tab” you should see on the tool-bar. That gives the [TEX]..[/TEX] wrap. Your LaTeX code goes between them.
Hello, Gurp925!
I got 8.57 and it was wrong.
First of all, you didn't round to the nearest kilogram.
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