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[MIMIC]

1) e^5x = 3^2x+1

As far as I can get:

ln e 5x = (2x+1) ln3
5x = (2x+1) ln3 (do I distribute and go from there?)

and

(pretend "@" is the pi symbol)

2) @^2x = 5^1-x

thanx.

 

[demi]

ADD X TO EACH SIDE

 

[fallout]

Judging by what you've done, I'm assuming you mean 3^(2x + 1) and not 3^2x + 1.

5x ln e = (2x+1) ln3
5x = (2x+1) ln3
5x = 2x ln 3 + ln 3
5x - 2xln 3 = ln 3
x (5 - 2 ln 3) = ln 3
x = ln 3 / (5 - 2 ln 3).

I'll spoiler the one below ... if you can get the first, you should be able to get this one. I'm also making another assumption about your lack of parentheses.

Spoiler

pi ^ 2x = 5 ^ (1 - x)
ln pi ^ 2x = ln 5 ^ (1 - x)
2x ln pi = (1 - x) ln 5
2x ln pi = ln 5 - x ln 5
2x ln pi - x ln 5 = ln 5
x (2 ln pi - ln 5) = ln 5
x = ln 5 / (x ln pi - ln 5).

 

[Dilbert]

One more point -- the laws of logarithms apply to all bases, not just e. Although natural log is the intuitive choice for problem #1, you could clean up problem #2 by choosing to take the log base 5 of each side in the first step.

 

[hobart]

It SEEMS as if you are on the natural log section of whatever math book yer rolling with. Word to the wise: just remember those LAWS by heart. Once you get them... you'll find it much easier to apply them to many of the problems you'll be seeing in the next week or so.

It also gives you an outlet for when you find yourself going: SHIT WHAT NOW?!?!

Your answer: Lemme try logs!

 

[Christoff Yurievich]

heh, we just started e^x today and finding the derivatives of those problems, so the natural log rule is very fresh.

 

[milanbaros]

I find it a little strange that I started and finished e^x today and I live in the UK.