How to solve it ?
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Solve the equation
2.3^x + 3^(1-x) = 11
Answer : x = 2.865
Thanks for the help.
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Originally posted by Snoopyies:
Solve the equation
2.3^x + 3^(1-x) = 11
Answer : x = 2.865
Thanks for the help.
Try applying log to the equation and you might be able to solve this.
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lg on both sides
lg [2.3^x + 3^(1-x)] = lg 11
lg [2.3^x + 3^(1-x)] = 1.0414
cannot solve further or 2.3^x + 3^(1-x) = 10^1.0414 = 11 back to the original.
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Originally posted by Snoopyies:
lg on both sides
lg [2.3^x + 3^(1-x)] = lg 11
lg [2.3^x + 3^(1-x)] = 1.0414
cannot solve further or 2.3^x + 3^(1-x) = 10^1.0414 = 11 back to the original.
lg [2.3^x + 3^(1-x)] = lg11
lg 2.3^x + lg 3^ (1-x) = lg 11
x lg 2.3 + (1-x) log 3 = lg 11
Try solving from here. Remember to apply your laws of log in which log x^a = a log x
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Originally posted by dkcx:
lg [2.3^x + 3^(1-x)] = lg11
lg 2.3^x + lg 3^ (1-x) = lg 11
x lg 2.3 + (1-x) log 3 = lg 11
Try solving from here. Remember to apply your laws of log in which log x^a = a log x
lg [2.3^x + 3^(1-x)] = lg11
lg 2.3^x + lg 3^ (1-x) = lg 11 ?
I think cannot multiply lg like that leh.
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Originally posted by dkcx:
lg [2.3^x + 3^(1-x)] = lg11
lg 2.3^x + lg 3^ (1-x) = lg 11
x lg 2.3 + (1-x) log 3 = lg 11
Try solving from here. Remember to apply your laws of log in which log x^a = a log x
Wrong liao leh. If so easy, my tutor will already post answer liao juz now.
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Originally posted by MasterMoogle:
Wrong liao leh. If so easy, my tutor will already post answer liao juz now.
Your tutor? I only had less than 4.5hrs of slp last night n already half aslp, will go think again tml if my brain gets working... Long time never play with log liao
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Originally posted by dkcx:
Your tutor? I only had less than 4.5hrs of slp last night n already half aslp, will go think again tml if my brain gets working... Long time never play with log liao
forbiddensinner la... I think he was online just now.
http://www75.wolframalpha.com/input/?i=(2.3)^x+%2B+3^(1-x)+%3D+11
look what I got! cannot calculate out one leh this question.
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I don't have a calculator:
x lg 2.3 + (1-x) lg 3 = lg 11
x lg 2.3 + 1 lg 3 - x lg 3 = lg 11
x lg 2.3 - x lg 3 = lg 11 - 1 lg 3
x (lg 2.3 - lg 3) = lg 11 - 1 lg 3
And then solve for x?
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Log has some multiplication or addition rule right?
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Originally posted by charlize:
I don't have a calculator:
x lg 2.3 + (1-x) lg 3 = lg 11
x lg 2.3 + 1 lg 3 - x lg 3 = lg 11
x lg 2.3 - x lg 3 = lg 11 - 1 lg 3
x (lg 2.3 - lg 3) = lg 11 - 1 lg 3
And then solve for x?
zzzzz cannot log like that der
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Originally posted by charlize:
Log has some multiplication or addition rule right?
lg ab = lg a + lg b
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I think better learn this question.
Might come out for the end of year exams.
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Originally posted by MasterMoogle:
forbiddensinner la... I think he was online just now.
http://www75.wolframalpha.com/input/?i=(2.3)^x+%2B+3^(1-x)+%3D+11
look what I got! cannot calculate out one leh this question.
But when substitute x = 2.865 into 2.3^x + 3^(1-x), the answer 11.00189664 is obtained.
So, x = 2.865 is the answer to the equation.
So, there should be some method to solve the equation, right ?
The problem is how to solve the equation to get the x answer of 2.865.
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Originally posted by charlize:
I think better learn this question.
Might come out for the end of year exams.
good to learn but no one give solution leh. i also waiting
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Originally posted by Snoopyies:
But when substitute x = 2.865 into 2.3^x + 3^(1-x), the answer 11.00189664 is obtained.
So, x = 2.865 is the answer to the equation.
So, there should be some method to solve the equation, right ?
The problem is how to solve the equation to get the x answer of 2.865.
i wish i know lor but like you i only sec school student. my 'shi fu' shen bing liao and now is dying on his bed so cannot help out lor
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This first step should be correct:
x lg 2.3 + (1-x) lg 3 = lg 11
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Originally posted by charlize:
This first step should be correct:
x lg 2.3 + (1-x) lg 3 = lg 11
smack u la tell u wrong liao u still say correct
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Originally posted by MasterMoogle:
smack u la tell u wrong liao u still say correct
Why wrong?
You bring down the power when you log the equation?
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Originally posted by charlize:
Why wrong?
You bring down the power when you log the equation?
no leh, u cannot log like this one lor in the 1st place
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log (a + b) =/= log a + log b
log (ab) then is log a + log b
looking for someone to do this correctly though, can't do it either =P
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Somebody who claims can't do simple maths is now expert at logarithm.
*Looking in the general direction of charlize*
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I did some googling.
Apparently log (a+b) is just log (a+b).
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Can google for me how to build a Faraday Cage?
I can't seem to find directions on how to construct one.
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Originally posted by Snoopyies:
Solve the equation
2.3^x + 3^(1-x) = 11
Answer : x = 2.865
Thanks for the help.
If this a H2 Maths question, we can use a graphic calculator
eg TI-84 Plus to solve it.
Steps
(1) Plot the curve y = 2.3^x
(2) Plot the curve y = 11 - 3^(1 - x)
(3) Find the x-ordinate of the intersecting point of the 2 curves
(a) press 2nd Calc
(b) choose 5 for intersect
(c) press enter 3 times
(4) x = 2.8647872