Math Is Fun Forum

Karimazer1

Guest

Rearranging Help!!!

Hey!

I have done a Q and can't seem to understand a very simple rearrangment.

So I have logR = y - (1/2)log(L1/L2)

So I raise each side to the power of 'e' since I want a solution for R... therefore I get

R = e^y ....

The answer given is R =e^y * ROOT(L2/L1) (or, (L2/L1)^0.5).

Why is this?

I understand the 'R' bit, and the e^y bit, but how does -(1/2)log(L1/L2) becomes (L2/L1)^0.5...why do they switch around? Why is this multiplied by e^y and where did the subtraction go?

I really can't do these kind of manipulations, help would really be great to allow me to progress!

Thank you!!!!

anonimnystefy

Real Member

From: Harlan's World

Registered: 2011-05-23

Posts: 16,049

Re: Rearranging Help!!!

Do you know the rule

a*log(b)=log(a^b) ? That is the on you can use with a=-1/2 and b=L1/L2.

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“Here lies the reader who will never open this book. He is forever dead.
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment
The knowledge of some things as a function of age is a delta function.

Karimazer1

Guest

Re: Rearranging Help!!!

Thank you so much for your reply.

I don't really understand that, I've found it online too whilst looking for log rules, and I still can't apply the transform to the result. Could you walk me through it?

Thank you... I thought I'd got the hang of such manipulations.

Bob

Administrator

Registered: 2010-06-20

Posts: 10,803

Re: Rearranging Help!!!

hi Karimazer1

Trouble with what you are saying is you'll have to do this:

So you really need to simpify the logs first:

Now you can do e^ on each side:

Then you can take the L terms over to the other side by inverting the L1/L2, and power 1/2 is the same as square root.

That'll give you the answer you want.

Bob

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Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

Karimazer1

Guest

Re: Rearranging Help!!!

Bob....

You ALWAYS come and save the day... I can't believe you wanted to work within the chemistry space when you have the patience of a saint! If only you were my math teacher...I probably would've solved these problems years ago!

Thank you so much...brilliant! I really understand it in those steps!

Bob

Administrator

Registered: 2010-06-20

Posts: 10,803

Re: Rearranging Help!!!

Thanks you for those kind words.  You're welcome.

Bob

---

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

anonimnystefy

Real Member

From: Harlan's World

Registered: 2011-05-23

Posts: 16,049

Re: Rearranging Help!!!

Ha! Saint! He's no Saint. He's better.

---

“Here lies the reader who will never open this book. He is forever dead.
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment
The knowledge of some things as a function of age is a delta function.