Look at the following and made some comments to assist.
Remember the log base is in [] brackets
Example 1
log [8] (x + 2) = 2 - log [8] 2
log [8] 1 + log [8] 2 = 2 / (x + 2)
Example 2
log [8] (x + 2) = 2 - log [8] 2
log [8] (x + 2) + log [8] 2 = 2
log (2x + 4) = 2
log (2x + 8) = 2
log 8
2x + 4 = 1.8
2x = 1.8 - 4
Example 3
log [8] (x + 2) = 2 - log [8] 2
log [8] x + log [8] 2 = 2 - log [8] 2
log [8] x + 2 log [8] 2 = 2
Example 4
log [8] (x + 2) = 2 - log [8] 2
log [8] (x + 2) = 1.67
log [8] 1.67 = x + 2
Example 5
log [8] (x + 2) = 2 - log [8] 2
log [8] (x + 2) + log [8] 2 = 2
log [8] 2(x + 2) = 2
log [8] (2x + 4) = 2
log [8] 2x = 2 - 4
Example 6
log [8] (x + 2) = 2 - log [8] 2
log (x + 2) / log 8 = 2 - log 2 / log 8
log (x + 2) / 0.903 = 2 - 0.33
log (x + 2) 0.903 * 1.667
log x + 2 = 1.5
log x = 1.5 - 2
log x = 0.5
x = 0.32
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EXAMPLE 1:
ReplyDeletelog [8](x+2)= 2-log [8]2
STEP 1: Bring all unknowns to the L.H.S. of the eq'n and, all constants to the R.H.S. of the eq'n
Rearranging yields: log [8](x+2) + log [8]2= 2
Since, log [b]a + log [b] c= log [b]ac
This implies, log [8](x+2) + log [8]2 = 2....now becomes...........
log [8] (x+2)(2)= 2
log [8] (2x+4) = 2
Since, log [b]a =c is equivalent to: b^c= a
This implies, 8^2= 2x+4
Therefore, 64 = 2x+4
64-4 = 2x
So, 60 = 2x
x = 30
Answer: x= 30
log [8] (x + 2) = 2 - log [8] 2
ReplyDeleteHey there!
I know this question may seem hard, but it really isn't!
It is however, a bit tricky!
First of all, you need to analyse the question paying attention to whether or not there exist a coefficient in the "log" function.
Then you look at the bases. Are they the same or are they different?
In this question, there exist no coefficients and the bases are the same, so it is safe to proceed!
log [8] (x + 2) = 2 - log [8] 2
In one solution we can group all the log terms. (i.e. bring all the logs on one side)
This will give:
log [8] (x + 2) + log [8] 2 = 2
RECALL: log a + log b = log (ab)
So we now have:
log [8] 2(x + 2)= 2 --------->(1)
Now, from the situation a^b = c, (1)is in the form {log [a] c = b}, where a is the base and b is the power.
From this form we can remove logs to show:
8^2 = 2(x + 2)
(i.e. a^b = c)
Then,
-> 2(x + 2) = 64
-> 2x + 4 = 64
-> 2x = 64 - 4
-> x = 60/2 = 30
EXAMPLE 1:
ReplyDeletelog [8]1 + log [8]2= 2/(x+2)
NOTE: THE LOG OF 1 TO ANY BASE IS ALWAYS 0, i.e.
log [a]1= 0.
So, the first term in the eq'n is 0.
Our eq'n now reads:
log [8]2= 2/(x+2)......cool?
Since, log [b]a= log a/log b
log [8]2 is in the form log [b]a, therefore:
log [8]2= log 2/log 8= 1/3......cool?
So, our eq'n now reads:
1/3= 2/(x+2)........CROSS MULTIPLYING YIELDS:
1*(x+2)= 2*3
(x+2)= 6
x = 6-2
x = 4
Ans.= 4
EXAMPLE 2:
ReplyDeletelog(2x+4)= 2.....when a sum appears to have NO
BASE, i.e. log x= y..(NO BASE), its base is understood to be 10.
So log (2x+4)= 2 is at base 10.
Therefore the eq'n now becomes:
log [10] (2x+4)= 2
So, 10^2 = 2x+4.....(from my first comment)
100 = 2x+4
100-4 = 2x
2x = 96
x = 96/2
x = 48
Ans.= 48
Nerd and Pheonex are the 2 students in my 100 students. Identify the error and why it is an error. Reason out why the person did what he did and explain an approach to avoid this error. Find the basic concept that is wrong, etc. Do not just give answers
ReplyDeletefor the example 1 i would bring all the logs to one side of the equal sign keeping in mind that (x+2)is one term and it cannot be separated from the log function. The approach to example one was half correct except that the 2 should not be divided by the (x+2).
ReplyDeletewhen you bring the logs to one side you will recognize that log[8]a + log[8]b is the same as log[8]ab so then you can bring the log[8](x+2)+ log[8]2 to log[8] (x+2)*2.
ReplyDeleteEXAMPLE 2:
ReplyDeletelog (2x+8)/log 8= 2
Therefore, log [10](2x+8)/ log [10]8= 2.......as you would understand by my second comment!
Now, CROSS MULTIPLYING results in:
log [10](2x+8)= 2*log [10]8
NOTE: log [10](2x+8)/log [10]8 IS NOT IN THE FORM, log [x](a/b)..........(as would result from log [x]a - log [x]b).....cool?
So log [10](2x+4)= 1.8062
So, 2x+4 = 10^1.8062.......(as you would understand by my first comment)
Therefore, 2x+4 = 64........(approx.)
2x = 64-4
2x = 60
x = 60/2
x = 30.
Ans.= 30
in e.g (2) the second line is completely correct; the person has grouped all the common terms together. in the third line the error begins! [8] is suppose to be after the log,however it was ommitted! in the forth line an algebric error has occured, the line suppose to be (2x+4) not (2x+8). the fifth line is correct execpt for one error which is that they have ommitted the log on the left hand side, therefore the answer is wrong!
ReplyDeletee.g (3) is totally wrong because brackets are never removed when a function is in front of it!
ReplyDeletein e.g 4 the second line is correct because the right hand side of the equation can be completely worked out on a calculator. the third line however is incorrect! the second line should have been put in exponential form and then x could of been found!
ReplyDeleteso it would of been:
log[8] (x+2) = 1.68
8^1.68 = x+2
32 = x+2
x = 30
e.g 5 would of been corrrect if not for the last line! as mentioned before brackets cannot be removed when a function is in frount of it!
ReplyDeleteeg of other functions are trigs
In question one ...the person wanted to put all the logs on one side..and the the so called known things on one side but by doin so in wrong manner..the removal of the brackets must neva happen until everything is worked out..
ReplyDeleteWhat it shud have been is log[8] (x+2)+ log[8] 2 = 2...where the logs on one side and the number(s) on the other...same bases on one side ..from here it is workable
ReplyDeleteFor example 1, first you have to group the like terms by putting logs on onside of the equation and all coefficient or variables on the other side of he equation. Now we evaluate the equation, whether or not the logs are being added or substracted. If added and the logs are of the same base multiple and if substracted we divide. From the previous steps the equation should now be in the form..log[2]8=3.. which can be switched to exponent form 2^3=8 to solve for the unknowns.
ReplyDeletelog [8](x+2)= 2-log [8]2
ReplyDeleteRearranging equation
log [8](x+2) + log [8]2= 2
log [8] (x+2)(2)= 2
log [8] (2x+4) = 2
8^2= 2x+4
64 = 2x+4
64-4 = 2x
60 = 2x
x = 30
Example 1
ReplyDeletelog [8] (x + 2) = 2 - log [8] 2
log [8] 1 + log [8] 2 = 2 / (x + 2)
I don't understand why you removed the log[8](x+2) and placed log [8]1 and divided by (x+2) on the R.H.S.
Example 2
ReplyDeletelog [8] (x + 2) = 2 - log [8] 2
place logs on one side of the = and the no. on the other side.
log [8] (x + 2) + log [8] 2 = 2
use the rule=> log[a]b +log[a]c = log[a]bc
LOG[8](X+2)(2)= 2KEEP THE BASE
log[8](2x + 4) = 2CHANG TO SAME BASE. DRAW LINE
log (2x + 4)/log 8 = 2CROSS MULTIPLY
LOG (2X + 4) = LOG8 * 2
LOG (2X + 4) = 1.8
USE: 2^3 = 8 <=> LOG[2]8 = 3
=> 10^1.8 = 2X + 4
63.1 =2x + 4
x = (63.1 - 4)/2
X = 29.55
Example 4
ReplyDeletelog [8] (x + 2) = 2 - log [8] 2
GET A COMMON BASE & DIVIDE IT.
LOG(X+2)/LOG8 = 2 -LOG2/LOG8
LOG(X+2)/0.9 = 2 - 0.33
LOG(X+2) = 1.67
CROSS MULTIPLY
LOG(X+2) = 1.67 * 0.9
LOG(X+2) = 1.5
USE: 2^3 = 8 <=> LOG[2]8 = 3
10^1.5 = X + 2
31.6 - 2 = X
X = 29.6
(log [8] (x + 2) = 1.67 THIS PART IS TOTALLY
log [8] 1.67 = x + 2) INCORRECT
NERD ALL THE EXAMPLES ARE THE SAME SO YOU CANNOT GET DIFFERENT ANSWERS. THE EXAMPLES JUST SHOWS DIFFERENT METHODS OF SOLVING THE SAME QUESTIONS.
ReplyDeletePLEASE TAKE THIS INTO CONSIDERATION & REDO