# Did I prove this correctly?

• Jan 27th 2014, 05:39 AM
sakonpure6
Did I prove this correctly?
Hi, there is this prove question that I did, but I am not sure If I did it correctly. I mean, am I aloud to manipulate both sides of the equation at the same time, and is it considered correct to end up with what I did?? Am I supposed to deal with this like an identity?

Attachment 30089Attachment 30090
• Jan 27th 2014, 05:54 AM
ebaines
Re: Did I prove this correctly?
It's hard to read the attachment, but it looks like the first line is

$\displaystyle h^2 + k^2 = 23 hk$

Please correct me if I misread that. Then on the second line you have written:

$\displaystyle \log(\frac {h+k} 5 ) = \frac 1 2 ( \log h + \log k)$

Not sure how you got that, but in general it's not true. Using the identity: $\displaystyle \log( \frac a b ) \ \log a - \log b$ in general:

$\displaystyle \log(\frac {h+k} 5 ) = \log(h+k) - \log 5$

It would help if can tell us what it is you are trying do with this problem.
• Jan 27th 2014, 05:54 AM
romsek
Re: Did I prove this correctly?
Quote:

Originally Posted by sakonpure6
Hi, there is this prove question that I did, but I am not sure If I did it correctly. I mean, am I aloud to manipulate both sides of the equation at the same time, and is it considered correct to end up with what I did?? Am I supposed to deal with this like an identity?

Attachment 30089Attachment 30090

you worked the problem backwards. You need to start with $\displaystyle h^2 + k^2 = 23 h k$ as in

$\displaystyle h^2 + k^2 = 23 h k \Rightarrow$

$\displaystyle (h+k)^2=25 h k \Rightarrow$

$\displaystyle \left(\frac{h+k}{5}\right)^2=hk \Rightarrow$

$\displaystyle 2\log\left(\frac{h+k}{5}\right)=\log(h)+\log(k) \Rightarrow$

$\displaystyle \log\left(\frac{h+k}{5}\right)=\frac{\log(h)+\log( k)}{2}$
• Jan 27th 2014, 06:13 AM
sakonpure6
Re: Did I prove this correctly?
If I worked backwards, is that still considered correct?
@ebaines I attached the question
• Jan 27th 2014, 08:45 AM
romsek
Re: Did I prove this correctly?
Quote:

Originally Posted by sakonpure6
If I worked backwards, is that still considered correct?
@ebaines I attached the question

not really but if you can work it backwards it's usually a trivial matter to just write it down in reverse order.
• Jan 27th 2014, 09:54 AM
ebaines
Re: Did I prove this correctly?
Quote:

Originally Posted by sakonpure6
If I worked backwards, is that still considered correct?@ebaines I attached the question

You didn't really do it backwards - that would have meant starting with $\displaystyle \log ( \frac {h+k}5 ) = \frac 1 2 (( \log h + \log k)$ and working from that to arrive at $\displaystyle h^2 + k^2 = 23 hk$. Your approach was a bit different - you started working backwards but mid-way threw in the initial given equation to ultimately arrive at 1 = 1, and hence conclude that the original statement must therefore be true. That works, but to avoid confusion like I had when I first saw your work you should notate your work to help the reader folllow your logic. The problem is that if you simply write a series of equations the reader assumes that each equation comes naturally from the one before it. But in your work it doesn't flow like that, so it's confusing. It's best to start with the given fact that $\displaystyle h^2 + k^2 = 23hk$ and from that proceed through a series of steps where each derives from the previous, to ultimately arrive at the desired expression.
• Jan 27th 2014, 10:05 AM
sakonpure6
Re: Did I prove this correctly?
Alright , thank you guys!