Another Logarithm!!! Quick please!! Watch

vector
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#1
Report Thread starter 12 years ago
#1
Hi,

Help with this please - ASAP!

25^x - 4*(5^x) + 3 = 0

Solution (x) between 0.5 and 1?

Thanks,
Michael
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insparato
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#2
Report 12 years ago
#2
25^x - 4*(5^x) + 3 = 0

rewrite this as

5^2x - 4(5^x) + 3 =0

let 5^x = y

so y^2 - 4y + 3 = 0

(y - 1)(y - 3)

y = 1 and 3

5^x = 1
5^x = 3

xlog5 = log 1
x = log1/log 5 = 0

xlog5 = log3
x = log3/log5 = 0.683 3 dp
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Chewwy
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#3
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= 5^x^2 - 4*5^x + 3 = 0 = (5^x-1)(5^x-3)

x = 0, ln3/ln5
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vector
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#4
Report Thread starter 12 years ago
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To the rescue again! Thanks for that man! You online tomorrow 7am - I'm bound to dream something up tonight lol!

Michael
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vector
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#5
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chewwy, i don't quite understand yours...
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insparato
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#6
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(Original post by vector771)
To the rescue again! Thanks for that man! You online tomorrow 7am - I'm bound to dream something up tonight lol!

Michael
I was online at 6am :P. I had a quick look round 7 didnt see anything lol.

Incidentally i dreamed about the exam last night lol like actually doing maths. I think that might be because the last 2 days ive been getting up at 8am and doing maths to 12am.
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Codefusion
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#7
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hey guys i just came back from the exam , anybody want an update ?
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Codefusion
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#8
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c1 or c2?
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