1. ## Logarithm question!!

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.

2. ## Re: Logarithm question!!

Originally Posted by Gurp925
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

2) $\log(36^{5x})=\log(6^{10x})~\&~\log(216^{x-7})=\log(6^{3x-21}).$

So $10x=3x-21$ solve.

3. ## Re: Logarithm question!!

oh my god i did that i cant believe i didnt factor the 3 into the 7 anyway thanks.

4. ## Re: Logarithm question!!

Originally Posted by Gurp925
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.
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".

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.
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)).

5. ## Re: Logarithm question!!

Hi HallsofIvy thanks for the response i get question #2 but i dont get one

i get up until this step logm=log(2.922)/.1964 after this step i am lost, i don't know how it turns into 2.922^1/0.1964

6. ## Re: Logarithm question!!

Originally Posted by HallsofIvy
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)).
It seems that the question is really:

$\log(36^{5x})=\log(216^{x-7}).$.

7. ## Re: Logarithm question!!

Originally Posted by Plato
It seems that the question is really:

$\log(36^{5x})=\log(216^{x-7}).$.
Yes this is right i just dont know to write in that format.

8. ## Re: Logarithm question!!

Originally Posted by Gurp925
Yes this is right i just dont know to write in that format.
It is LaTeX coding.

[TEX]\log(36^{5x})=\log(216^{x-7}).[/TEX] gives $\log(36^{5x})=\log(216^{x-7}).$.

9. ## Re: Logarithm question!!

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.

anyone ?

11. ## Re: Logarithm question!!

Originally Posted by Gurp925
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.
I presume you know that "a log(b)= log(b^a)" log(2.922)/.1924= (1/1.924) log(2.922)= log(2.944^(1/.1924).

And, of course, if log(a)= log(b) then a= b (because log is "one to one") so log(m)= log(2.922)/.1964= log(2.922^(1/.1964)) becomes m= 2.922^(1/.1964)

12. ## Re: Logarithm question!!

Originally Posted by Gurp925
, 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 $\boxed{\Sigma}$ on the tool-bar. That gives the [TEX]..[/TEX] wrap. Your LaTeX code goes between them.

Thanks all!

14. ## Re: Logarithm question!!

$\log(2^{3x})$

TEST

15. ## Re: Logarithm question!!

Hello, Gurp925!

$(1)\;\;\log E\:=\:\log(10.61)+0.1964\log(m)$

$\text{If }E = 31,\,\text{ solve for }m.\:\text{ Round to nearest kilogram.}$

I got 8.57 and it was wrong.

First of all, you didn't round to the nearest kilogram.

We have: . $\log(10.61) + 0.1964\log(m) \;=\;\log(31)$

. . . . . . . . . . . . . . . . . $0.1964\log(m) \;=\;\log(31) - \log(10.61)$

. . . . . . . . . . . . . . . . . $0.1964\log(m) \;=\;\log\left(\frac{31}{10.61}\right)$

n . . . . . . . . . . . . . . . . . . . . $\log(m) \;=\;\frac{1}{0.1864}\log\left(\frac{31}{10.61} \right)$

. . . . . . . . . . . . . . . . . . . . . $\log(m) \;=\;\log\left(\frac{31}{10.61}\right)^{\frac{1}{0 .1964}}$

. . . . . . . . . . . . . . . . . . . . . . . . $m \;=\;\left(\frac{31}{10.61}\right)^{\frac{1}{0.129 64}}$

Therefore: . $m \;=\;234.9134547 \;\approx\;235$

$(2)\;\;\log(36^{5x}) \:=\: \log(216^{x-7})$

We have: . $\log(6^2)^{5x} \;=\;\log(6^3)^{x-7}$

. . . . . . . . $\log\left(6^{10x}\right) \;=\;\log\left(6^{3x-21}\right)$

Therefore: . . . . $10x \:=\:3x-21$

n . . . . . . . . . . . $7x \;=\;-21$

. . . . . . . . . . . . . $x \;=\;-3$

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