first we check for coefficients there is none we notice that the bases are the same so log[7](x^2 = log[7](2x-1) ~ log(x^2) = log(2x-1) since this is so we can remove the logs and we get a quadratic equation x^2 = 2x-1 ~ x^2 - 2 + 1
Question 2 Following the rule given by goldfinger in problem3.....you can rewrite the equtaion for question2--->2log[3]x=log[3](x-1)(4).... since log is to the same base in both sides of the equation....it can be cancelled..... leaving----->2x=(x-1)(4) which then leads to---->2x=4x-4..... which can be solved---->2x-4x=4-----> -2x=4---->x=-2
Hey 'original_dapper17' i think you made a mistake. you didnt remove your cofficient, so your answer is wrong, look i did it over so cheak for you error:
1st thing to do is to check if there is any coefficients:
log[3]x^2 = log[3]4 + log[3] (x-1)
we know from our log rules that when adding common logs we multiply:
wel as mysticwings said when adding to logs with the same base you multiply the powers therefore you will get. log2^(x^2-2x)=3 now using 2^3=8 = log2^8=3
i think you will get 2^3=x^2-2x 8=x^2-2x thats as far as i can go.
goldfinger, i like how u explained question one compared to mysticwings, ur steps in working it was clear, mysticwings...thats something 4 u to work on...
firstly remove the 2 and bring it as a power of x.Then we work out the right hand side. when adding logs with the same base we multiply so therefore we get
log[3]x^2 = log[3]4(x-1)
then we could remove the logs on both side and we get a quadratic expression
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problem 1
ReplyDeletefirst we check for coefficients
there is none
we notice that the bases are the same
so log[7](x^2 = log[7](2x-1) ~ log(x^2) = log(2x-1)
since this is so
we can remove the logs
and we get a quadratic equation
x^2 = 2x-1 ~ x^2 - 2 + 1
factorize we get
(x-1)(x-1)
so x = 1
problem 3
ReplyDeleteusing rule log[b]a + log[b]c ~ log[b]ac
therefore log[2]x + log[2](x-2) = log[2](x)(x-2)
equation ~ log[2](x^2-2x)= 3
using log rule log[a]b = logb/loga
(x^2-2x)/log2 = 3
so (x^2-2x) = 3log2
work out the numbers
and you get a quadratic equation
use the quadratic equation
(-b+-√b^2-4ac)/2a
Question 2
ReplyDeleteFollowing the rule given by goldfinger in problem3.....you can rewrite the equtaion for question2--->2log[3]x=log[3](x-1)(4)....
since log is to the same base in both sides of the equation....it can be cancelled.....
leaving----->2x=(x-1)(4) which then leads to---->2x=4x-4.....
which can be solved---->2x-4x=4-----> -2x=4---->x=-2
no 1
ReplyDeletesince we have common logs we can drop logs on both sides:
x^2 = 2x - 1
from here its just normal algebra
x^2 - 2x + 1 = 0
(x-1)(x-1) = 0
x = 1
no 2:
ReplyDeleteHey 'original_dapper17' i think you made a mistake. you didnt remove your cofficient, so your answer is wrong, look i did it over so cheak for you error:
1st thing to do is to check if there is any coefficients:
log[3]x^2 = log[3]4 + log[3] (x-1)
we know from our log rules that when adding common logs we multiply:
log[3]x^2 = log[3]4(x-1)
now we can drop logs on both sides:
x^2 = 4(x-1)
from here its just algebra
x^2 = 4x - 4
x^2 - 4x + 4 = 0
(x-2)(x-2)
x = 2
no 3
ReplyDeletewhen adding common logs we know we must multiply:
log[2]x(x - 2)= 3
log[2] (x^2 - 2x) = 3
now we can put this in exponent form:
2^3 = x^2 - 2x
8 = x^2 - 2x
x^2 - 2x -8 = 0
(x-4)(x+2)
x = 4 and x = -2
Solve log2(x) + log2(x – 2) = 3
ReplyDeletewel as mysticwings said when adding to logs with the same base you multiply the powers therefore you will get.
log2^(x^2-2x)=3
now using 2^3=8 = log2^8=3
i think you will get 2^3=x^2-2x
8=x^2-2x
thats as far as i can go.
goldfinger, i like how u explained question one compared to mysticwings, ur steps in working it was clear, mysticwings...thats something 4 u to work on...
ReplyDeletethe other 2 question giving a little problem to comprehend, i think ill go see miss, logs are not my strong point.
ReplyDeleteI agree with hellsream it is good to show all your steps, it can spare you from losting alot marks, i am sure.
ReplyDeleteQuestion 1
ReplyDeleteSolve log[7](x^2) = log[7](2x – 1).
since there are common logs we can drop logs on both sides:
x^2 = 2x - 1
x^2 - 2x + 1 = 0
Solve using quardratics
x-1)(x-1) = 0
x = 1
Question 2
ReplyDeleteSolve 2log[3](x) = log[3](4) + log[3](x – 1)
We can multiply on the left hand side of the equation since the both logs have the same base.
log[3]x^2 = log[3]4(x-1)
logs on both sides since it have the same base:
x^2 = 4(x-1)
x^2 = 4x - 4
x^2 - 4x + 4 = 0
Solve using quardratic expression
(x-2)(x-2)
x = 2
Question 3
ReplyDeleteSolve log2(x) + log2(x – 2) = 3
Since both logs have the same base, we can multiply:
log2 x(x-2) = 3
log2 (x^2 - 2x) = 3
This can now be switched to exponential form:
2^3 = x^2 - 2x
8 = x^2 - 2x
Solve using quadratic expression
x^2 - 2x -8 = 0
(x-4)(x+2)
x = 4 and x = -2
Question 1.
ReplyDeletelog[7](x^2) = log[7](2x – 1)
the logs have the same bases so we could remove the logs and we get a Quadratic expression
x^2 = 2x – 1
Then we solve the expression
x^2 - 2x – 1 = 0
(x-1)(x-1) = 0
x = 1
Question 2
ReplyDelete2log[3](x) = log[3](4) + log[3](x – 1)
firstly remove the 2 and bring it as a power of x.Then we work out the right hand side. when adding logs with the same base we multiply so therefore we get
log[3]x^2 = log[3]4(x-1)
then we could remove the logs on both side and we get a quadratic expression
x^2 = 4(x-1)
x^2 = 4x - 4
x^2 - 4x + 4 = 0
(x-2)(x-2) = 0
x = 2
Question 3
ReplyDeletelog2(x) + log2(x – 2) = 3
the logs have the same base so we multiply therefore we get
log2 x(x-2) = 3
log2 (x^2 - 2x) = 3
Now put the log to exponent we get
2^3 = x^2 - 2x
8 = x^2 - 2x
now we can solve the quadratic expression
x^2 - 2x -8 = 0
(x-4)(x+2)
x = 4 , x = -2
my..question...and
ReplyDeletesolution
log[2]P=log[2]23.9+0.450
log[2]P-log[2]23.9=0.45
log[2]P/23.9=0.45
2^0.45=P/23.9
P=1.37*23.9=32.6
answer=32.6
another...question..with..solution
ReplyDeletelogx^2-1=log(1-2x)
logx^2-log10=log(1-2x)
logx^2/10=log(1-2x)
x^2=10-20x
0=x^2+20x-10
question..solution
ReplyDelete2log[2]3-log[2]x=log[2]45
log[2]9-log[2]45=log[2]X
log[2]9/45=log[2]X
x=9/45
question...solution
ReplyDelete2log2+logX=log3
log2^2+logX=log3
logX=log3-log4
logX=log3/4
X=3/4
remember:
ReplyDelete2^3=8<==>log[2]8=3
log[7]1=0
7^0=1
log[5]5=1
5^1=5
log[2]ab=log[2]a+log[2]b
log[2]b/a=log[2]b-log[2]a
log[2]a^b=blog[2]^a
Ques1.Solve log[7](x^2) = log[7](2x – 1).
ReplyDeletex^2 = 2x - 1
x^2 - 2x – 1 = 0
(x-1)(x-1) = 0
x-1=0
x=1
well this is how i think to wrk this problem
2log[3](x) = log[3](4) + log[3](x – 1)
ReplyDeletelog[3]x^2 = log[3]4(x-1)
x^2 = 4(x-1)
x^2 = 4x - 4
x^2 - 4x + 4 = 0
(x-2)(x-2) = 0
x = 2
1)log[7](x^2) = log[7](2x – 1)
ReplyDeletesince the logs is the same we could cancel the out
x^2 = 2x – 1
put it equal to zero
and solve
x^2 - 2x – 1 = 0
(x-1)(x-1) = 0
3)log2(x) + log2(x – 2) = 3
ReplyDeleteapply the rules when adding logs with the same base. we multiply
log2 (x^2 - 2x) = 3
now we can put in exponent form
2^3=8 -----> log2^8=3
2^3 = x^2 - 2x
8 = x^2 - 2x
solve the quadratic expression
x^2 - 2x -8 = 0
2) 2log[3](x) = log[3](4) + log[3](x – 1)
ReplyDeleteremove all coefficient
log[3](x)^2 = log[3](4) + log[3](x – 1)
when adding logs apply the rule. so we multply
log[3]x^2 = log[3]4(x-1)
we cancel the logs
x^2 = 4(x-1)
x^2 = 4x - 4
x^2 - 4x + 4 = 0
solve quadratic