Below show some exponent approaches. This time the power will be in brackets i.e. ( )
Example 1
5 (2x - 1) = 6 (x + 3)
Multiply left by 6
Multiply right by 5
30 (2x - 1) = 30 (x + 3)
2x -1 = x + 3
Example 2
5 (2x - 1) = 6 (x +3)
log (5) 2x - 1 = log (6) x + 3
log 2x - 1 / log 5 = log x + 3 / log 6
12x - 6 = 5x + 15
Example 3
5 (2x - 1) = 6 (x +3)
log (5) 2x - 1 = log (6) x + 3
2x - 1 log 5 = x + 3 log 6
2x - 1 log 5 - x + 3 log 6 = 0
log [5] (2x - 1) + log [6] (-x + 3) = 0
log 10x - 5 + 9 - x = 0
log 9x + 9 = 0
log 9x = 9
36 comments:
in example 2 wouldn't it be this:-
log (5) (2x-1)= log (6) (x+3)
log (5) 6 (x+3)= 2x-1
draw a line
log 6 (x+3) =log 2x-1
lo5 1
drop logs on both side and cross multiple
(x+3)6 = (2x-1)5
6x+18 =10x-5
10x-6x=18+5
4x=23
x= 23/4
= 5.75
well for example one, when you multiply the left side by the coeffecient of the brackets from the right side and you multiply the right side by the coeffecient of the brackets from the left side, you will get the same coeffecient on both the left and right sides of the brackets to be the same, 30, thus cancelling it out and then you solve.
Example2 5(2x-1)=6(x+3) should this be Log 6(x+3)=(2x-1)
5
when you converting to logs????
i dont understand how logs came into that.... ent 5(2x-1) = 6(x+3).. all u suppose to do is expand and find x.. it will be 10x-5= 6x+18.... = 10x-6x= 18+5...= 4x=23...x=23/4??? wouldnt that be it??
cool face you forgot to read what miss said to the top. the number inside the bracket symbolises the powers. so its not as simple as expanding! its logs!
miss i could see the mistakes in example 1 and example 2 but what really went on wiht example 3. i have no idea!
answer= take logs on both sides of the equation log 5 (2x+1) = log 6 (x+3)
2x+1 log 5 = x+3 log 6
2x+1/X+3 = log 6/log 5
and solve
Can someone please tell me what is going on in here cause i am really lost what am i suppose do am i suppose to identify a mistake or am i suppose to just say something about the examples in my own words?????????
I don quite understand these methods at all I am quite puzzled??*//*/??
after working this out the way I know about this I got a totally different answer. Are these methods correct at all???
looking at example 1..the method used is correct. i would have done the same by tryn to bring all the bases to the same value, then solving. with respect to example 2 the third line is confusing. i tink its wrong. Wat do u tink??? i dont kno about dat method. please sum 1 explain??
the last example also confuses me. 5 (2x - 1) = 6 (x +3)
log (5) 2x - 1 = log (6) x + 3
2x - 1 log 5 = x + 3 log 6
i fully understand up to this point. from here i wud have worked out the values of log 5 and log 6 on my calculator, and substitute the values into the equation and then solve for x. i tink my method makes more sence. wat u tink??
bootz make sure and read this you are to make any corrections on these questions to help those who struggle with it. or is it that you struggling too
In example 1 if elements in the brackets are exponents it cannot just become a base. There is still a value for the base although it has been simplified, that value is 1.
finishing the answer
2x+1/x+3 = 1.113
x+3 = 2x+1(1.113)
x+3 = 2.23x + 1.113
group x terms
2.23x -x= 3-1.113
x = 1.53
TADA!!!!!!!!!!! there is your answer
i agree with cokebaby in saying the method used in example 1 is correct. if the base of two or more exponents are the same then the bases can dropped leaving only the powers behind and so the value of x can be found
example 1 .... wrong wrong and wrong jed. you cannot multiply a power by a base like that. somebody please help this situation
hey people waz up i am a little confuse about example 3 could someone explain it to me in better terms
dark angel!!! deres nothing to explain.. it jus wrong. the first 4lines make sum sense.. read my comment on the last example.. maybe ul understand!!!
wong fei hong!!! wat sooo wrong about example 1???? it looks correct, but im abit unsure cuz i neva used dat method to solve a problem like dat...
boots u ought to do both
N2o...ur first comment comment confuses me..i tink u have the wrong idea. your first line, where u took logs on both sides was correct.
5 (2x - 1) = 6 (x +3)
log (5) 2x - 1 = log (6) x + 3.
y dont u work out the value of log 5 and log 6 on ur calculator and just solve for x from there..???
In example 1. It is not possible to multiply both sides by different numbers. U must multiply both sides by the same the same number.
this debate about example one is making me even more confused could some one explain what is so wrong about it
Example 1 is totally incorrect. You cannot multiply bases like you multiply ordinary numbers or expressions since the value of the base and power are linked.
TO n2o:
Dont quite get what you are saying you are confusing me.Someone help me get this?
Well sleepy my friend you are correct again that would have been my approach.
Yea ichigo,what or how logs came into this example,explain.
5(2x-1)=6(x+3)
log5(2x-1)=log6(x+3)
Using rules
2x-1log5 =x+3log6
2x-1/x+3=log6/log5
Using calculator
2x-1/x+3=0.778/0.699
Now cross-multiply
0.699(2x-1)=0.778(x+3)
1.398x-0.699=0.778x+2.334
1.398x-0.778x=2.334+0.699
0.62x=3.033
x=3.033/0.62
x=4.89
Tell me if im in the dark,meaning am i wrong......
To cool-face:
I think that the logs came into the equation to maintain a balance on both sides of the equation because as we know in maths whatever we do on one side of an equation it must also be done on the other.
Then as we do this we can solve the equation.
And I hope that it helps you to understand how logs came into the equation.
I am seeing errors in all three examples, for instance in the first example logs weas not taken at all and also none of the examples made x the subject of the equation.
AM I WRONG? CORRECT ME IF I AM ANYONE.
TO cokebaby:
I don't understand how you are seeing the first example to be looking correct.
If the left is multiplied 6 and the right by 5 then there is no balance in the equation because it is multiplied by two different numbers.
I may be wrong, enlighten me.
To everyone:
Are any of the examples listed by miss fariel correct.If there is/are any correct please bring them to my attention because I have seen errors in all three examples.
I am a bit confused!
i think all three examples are wrong. in example one both sides of the equation are multiplied by different numbers therefore it is not balanced remember that whatever is done on one side of the equation must be done on the other side so they will still be equal. example 2 is wrong since when logs are applied to both sides it means log of the entire expression not each individual variable or number and it is not possible to divide like that according to the log laws. i understand example 3 up until the fifth line i don't think the rest is correct and it can't really be solved in this format. i'm a bit confused as in how to solve this problem moose way seems to be correct however when the value for x is substituded back into the equation it could never be correct they are still not equal.
could someone explain please!
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