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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!

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

thanks!

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#3

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.

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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#5

(Original post by

2xlog3 = 2-xlog4

2x/2-x = log4/log3

but then thats difficult to find x from there

**madhapper**)2xlog3 = 2-xlog4

2x/2-x = log4/log3

but then thats difficult to find x from there

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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(Original post by

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])

**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])

???????????????

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#7

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

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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#8

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)

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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#9

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

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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#11

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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#12

(Original post by

solve 3^(2x)=4^(2-x)

**sarah12345**)solve 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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