How to solve it ?

Snoopyies SG

Solve the equation

2.3^^x + 3^^(1-x) = 11

Answer : x = 2.865

Thanks for the help.

dkcx SG

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.

Snoopyies SG

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.

dkcx SG

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

Snoopyies SG

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.

MasterMoogle SG

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.

dkcx SG

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

MasterMoogle SG

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.

charlize SG

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?

charlize SG

Log has some multiplication or addition rule right?

MasterMoogle SG

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

MasterMoogle SG

Log has some multiplication or addition rule right?

lg ab = lg a + lg b

charlize SG

I think better learn this question.

Might come out for the end of year exams.

Snoopyies SG

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.

MasterMoogle SG

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

MasterMoogle SG

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

charlize SG

This first step should be correct:

x lg 2.3 + (1-x) lg 3 = lg 11

MasterMoogle SG

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

charlize SG

smack u la tell u wrong liao u still say correct

Why wrong?

You bring down the power when you log the equation?

MasterMoogle SG

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