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sarah12345
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#1
Report Thread starter 15 years ago
#1
ok i think im bein abit thick but i keep gettin the wrong answer

solve the equation 3^2x=4^2-x

thanks!
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madhapper
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#2
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#2
do u mean 4^2 - x or 4^(2-x)?
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nas7232
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#3
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ok i think im bein abit thick but i keep gettin the wrong answer

solve the equation 3^2x=4^2-x

thanks!

2xlog3 = 2-xlog4
2xlog3 - (2-xlog4) = 0
sry im tired i can't think

factorise that by taking x as a common letter you will get x( something something) Thus you can solve for x, i think that's how you do it. I haven't completed my c2 logs bit yet.
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madhapper
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#4
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2xlog3 = 2-xlog4
2x/2-x = log4/log3

but then thats difficult to find x from there
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manps
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#5
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(Original post by madhapper)
2xlog3 = 2-xlog4
2x/2-x = log4/log3

but then thats difficult to find x from there
if thats true then
log4/log3 = 1.26...

2x/(2-x) = 1.26...
2x = (2-x)1.26
2x = 2.52...-1.26..x
2x - 1.26x = 2.52
0.73x = 2.52..
x = 2.52/0.73
x = 3.419022...

or
x = 2(log4/log3)/(2 - [log4/log3])
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sarah12345
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Report Thread starter 15 years ago
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(Original post by manps)
if thats true then
log4/log3 = 1.26...

2x/(2-x) = 1.26...
2x = (2-x)1.26
2x = 2.52...-1.26..x
2x - 1.26x = 2.52
0.73x = 2.52..
x = 2.52/0.73
x = 3.419022...

or
x = 2(log4/log3)/(2 - [log4/log3])
the answer is 0.774
???????????????
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nas7232
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#7
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try my method :P
another method..
3^2x=4^2-x

take logs to the base 3 of each side
you will get [logs are to the base 3 here]
2x = (2-x)log4
2x / (2-x) = log 4

or take logs to the base 4
2xlog3 = 2-x
2-x / 2x = log 9 + log 3x
Therefore: 6x^2 - (2-x) = -log9

6x^2 - 2 - x + log 9 = 0

I give up
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manps
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#8
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i got ur answer

3^2x = 4^(2-x)
2x.log3 = (2-x).log4
log3/log4 = (2-x)/(2x)
0.79248.... = (2-x)/(2x)
1.584962501x = 2-x
2.584962501x = 2
x = 2/2.584962501
x = 0.774 (3dp)
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kikzen
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#9
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3^2x=4^2-x
2xln3=(2-x)ln4
2xln3+xln4=2ln4
x(2ln3+ln4)=2ln4
x = 2ln4/(2ln3+ln4)

but i dont have a calculator to check that
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john !!
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nas you want to take logs to the base 10 usually, so your calculator can do it
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nas7232
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#11
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v.true. However, i didnt think sarah liked my method with lg, log to the base 3 and 4 makes gets rid of the nasty bits of the equation. Just thinking maybe something may come out of it which would solve the question.. unlucky though
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username9816
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#12
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(Original post by sarah12345)
solve 3^(2x)=4^(2-x)
3^(2x) = 4^(2-x)
---> 2xln3 = (2-x)ln4
---> 2x/(2-x) = ln4/ln3
---> (2xln3)/(ln4) = 2-x
---> x(2ln3/ln4) = 2-x
---> (2-x)/x = ln9/ln4
---> 2/x - 1 = ln9/ln4
---> 2/x = (ln9/ln4) + 1
---> x/2 = 1/[(ln9/ln4) + 1]
---> x = 2/[(ln9/ln4) + 1]
---> x = 0.774 (3.S.F)
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