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 July 7th, 2013, 12:53 PM #1 Senior Member   Joined: Apr 2013 From: Ramallah, Palestine Posts: 349 Thanks: 0 Check my work please. Thank you Solve each equation for x. $e^{7-4x}= 6$ Here is my work. $e^{7-4x} -6= 0$ $\ln(e^{7-4x})\ln(-6)= 0$ $7 - 4x - \ln(6)= 0$ add + 4x $7 - \ln(6)= 4x$ $\dfrac{7 - \ln(6)}{4}= \dfrac{4x}{4}$ $\dfrac{7 - \ln(6)}{4}= x$ but the book is saying the answer is $\dfrac{1}{4} (7 - \ln 6)$ <-- notice they don't put the 6 in parentheses like I did ln(6) is there any significance in this? or no is 1/4 * somethinig the same as something / by 4 ? Sorry if this sounds like an elementary type question. I just want to understand if my answer is correct or not?
 July 7th, 2013, 01:29 PM #2 Senior Member   Joined: Jun 2013 From: London, England Posts: 1,316 Thanks: 116 Re: Check my work please. Thank you I'm amazed you can manipulate exponentials and logs and not recognise that your answer is the same as the book answer. You made things more difficult for yourself by bringing the 6 across to the LHS. You should have left it where it was! After that, your working goes off the rails a bit. The third line doesn't follow from the second. I'd leave the 6 on the RHS and try again!
 July 7th, 2013, 01:37 PM #3 Senior Member   Joined: Apr 2013 From: Ramallah, Palestine Posts: 349 Thanks: 0 Re: Check my work please. Thank you Yea Pero, well I'm glad your a genius. When things don't look the same, I get confused. Like for example, why are they not saying ln(6) in the book? They are just saying ln 6. There has got to be a difference, like is it LN OF 6 or is it LN * 6?
July 8th, 2013, 04:44 AM   #4
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Quote:
 Originally Posted by Shamieh $\ln(e^{7-4x})\ln(-6)= 0$
That should have been $\ln(e^{7-4x})\,-\,\ln(6)\,=\,0.$

Strictly speaking, one should minimize ambiguity by writing ln(6) rather than ln 6, but the rules are often relaxed where the intention is reasonably clear. For example, cos 2A is assumed to mean cos(2A). When using LaTeX, you should type \ln instead of ln, so that the function name is recognized and not rendered in italics. I amended your post to do this for you.

 July 8th, 2013, 11:57 AM #5 Senior Member   Joined: Apr 2013 From: Ramallah, Palestine Posts: 349 Thanks: 0 Re: Check my work please. Thank you Ahh, cool I understand now Skipjack. Thank you, that was the answer I was looking for.
July 8th, 2013, 08:19 PM   #6
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Re: Check my work please. Thank you

Quote:
 Originally Posted by Shamieh $e^{7-4x}= 6$
REMEMBER: if a^p = b, then p = log(b) / log(a)
So:
7 - 4x = log(6) ; almost finished!

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