Don't know the first thing about this one: 24=e^(x+1)

SyrenaV

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Aug 19, 2007
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4
Fixed the math problem. Now it should make sense.

Yeah. I haven't got the first clue as to what I'm supposed to with this, and it's part of my AP Calculus summer work. The part that's getting me is the e (not sure if it's the actual number like in a natural log or what), and the function in the exponent. Any suggestions?

24=e^x+1
 
OH! Sorry.

I got my keyboarding wrong both times. It's SUPPOSED to say 24=e^x-1. Oops.....
 
SyrenaV said:
I got my keyboarding wrong both times. It's SUPPOSED to say 24=e^x-1. Oops.....
Okay, so the equation is "24 = e<sup>x</sup> - 1". But what are the instructions? What are you supposed to do with this equation? :oops:

Please be specific, showing all of your work so far. Thank you! :D

Eliz.

P.S. To learn about the natural exponential "e", try reading some of the lessons available online. :wink:
 
Okay, I really shouldn't have made this topic at ten thirty after trying for hours to get this problem... :oops:

The equation reads: 24 = e<sup>x-1</sup>. My job is to solve for x.

I'm pretty sure I'm just having letter overload, but I think I'm supposed to get the ln of 24 and then set that equal to (x-1). (It's been a while since I've done this (like, maybe two years), so I really need to refresh my brain on logs and exponentials.)

ln(24)=x-1

After I get the ln of 24, do I add 1 to get x by itself?
 
SyrenaV said:
After I get the ln of 24, do I add 1 to get x by itself?
Yes, though the expected answer is probably "ln(24) + 1", not the decimal approximation. :wink:

Eliz.
 
stapel said:
SyrenaV said:
After I get the ln of 24, do I add 1 to get x by itself?
Yes, though the expected answer is probably "ln(24) + 1", not the decimal approximation. :wink:

Eliz.

Thank you! :mrgreen: This was one of those things that I know, but it's so far back in my subconscious, I can't ever get to it without forcing my brain out of its usual state. I was terrified of showing up in AP Calculus with NO IDEA what I was doing.
 
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