1. ## Transposition....

Hi I am new to this forum and require some help on transposition.... B is required to be the subject and the formula is

C = B x log28/ 1 + A

Any help would be great

2. ## re: Transposition....

Originally Posted by red9
Hi I am new to this forum and require some help on transposition.... B is required to be the subject and the formula is

C = B x log28/ 1 + A

Any help would be great
if you meant ...

$C = B \cdot \frac{\log_2{8}}{1+A}$

$C = B \cdot \frac{3}{1+A}$

$\frac{C(1+A)}{3} = B$

3. ## re: Transposition....

Originally Posted by red9
Hi I am new to this forum and require some help on transposition.... B is required to be the subject and the formula is

C = B x log[SIZE=1]2[SIZE=3]8/ 1 + A
Why don't you use grouping symbols.
Do you mean $C=\frac{Bx\log_2(8)}{1+A}$ OR $C=\frac{Bx\log_2(8)}{1}+A~?$

Or something altogether different?

4. ## re: Transposition....

Hi Skeeter thanks very much, transposition is bit of a downfall for me but Im sure if I keep trying it will sink in...
Thanks again

5. ## Re: Transposition....

Originally Posted by red9
C = B x log[SIZE=1]2[SIZE=3]8/ 1 + A
Please use * as multiplication sign; also, brackets required: (1 + A)

As a suggestion, seems to be nothing preventing simplification:

Let k = log(2)8; your equation becomes:
C = B*k / (1 + A)
cross multiply:
Bk = C (1 + A)
divide by k:
B = C(1 + A) / k

6. ## Re: Transposition....

Thank you Wilmer,
How about to get A as te subject from the orignal equation?

7. ## Re: Transposition....

NOTE: edited my previous post...had a "watching hockey game" error!
Originally Posted by red9
How about to get A as te subject from the orignal equation?
original: C = B*k / (1 + A)
cross multiply:
C(1 + A) = Bk
C + AC = Bk
AC = Bk - C
A = (Bk - C) / C

I'll trust that you're "following/understanding" this, not just using as homework answer...

If you come back with a similar problem......SHOW your work, else I'm busy!