maths : logarithms !! (2 Viewers)

[sle3pe3bumz]

well ive got a little problem with logarithms in maths .. i can seem to know what to do first ..
heres the question that im on atm ..

log(5x + 12) - log(2x - 3) = 2

would ii like bring the log out and (5x + 12) / (2x - 3) or expand or what ? .. its confusing me ! ><" .. & ii hate logs !!

 

[SoulSearcher]

What base is the log?

Generally, you would go:

log_a(5x + 12) - log(2x - 3) = 2
log_a{(5x+12)/(2x-3)} = 2
(5x+12)/(2x-3) = a²

From there on, you would solve for x, but it would be easier if we knew what base the logarithm was.

Eventaully, x would be
-(3a²+12)/(5-2a²)

 

[sle3pe3bumz]

well the base would probably be 10 ..
& how the hell did you get -(3a2+12)/(5-2a2)? .. ahaha .. ><" .. damnnnn ..

 

[SoulSearcher]

If it is 10, then x would be -(3*100+12)/(5-2*100) = -312/-195 = 8/5 = 1.6

Ok, (5x+12)/(2x-3) = a²
5x+12 = a²(2x-3)
5x+12 = 2xa²-3a²
5x-2xa² = -3a²-12
x(5-2a²) = -(3a²+12)
x = -(3a²+12)/(5-2a²)

 

[sle3pe3bumz]

oh wow ! .. seems so simple .. thank you soulsearcher .. =]

 

[SoulSearcher]

No problem You'll get used to those kinds of questions after you do a few of them :uhhuh:

 

[sle3pe3bumz]

oh okay .. ahah .. hope ii get used to em .. ><" .. anyways .. thanks again .. =]

 

[z600]

find x right? and the base in 10

log(5x+12)-log(2x-3)=2

log (5x+12)/(2x-3)=2 change into fractions

10^2= (5x+12/2x-3) change into index law

100= (5x+12/2x-3) 10^2=100

200x-300=5x+12 cross multiply

195x=312 solve

x=1.6

some log questions get nasty

 

[sle3pe3bumz]

ummm .. ive got another few questions which ii would like help with .. ><"

log(x^2 - 9) - log(x - 5) = log(x - 1/3)


log *base 2* (2^(2x) - 56) - x = 0


5^(2x) - 5^(x+1) = -4


ii still dont what to do first !! ><" ..

 

[ice ken]

is that 3 different questions?

 

[sle3pe3bumz]

lols yess .. ><"

 

[ice ken]

lols yess .. ><"

these questions are helllll hard i forget alll my log rules omg. killing my brain

are u suppose to find x rite?

 

[ice ken]

only can do 1st one.

log (x^2-9)/log(x-5) = log (x-1/3)
(x^2-9)/(x-5)= (x-1/3)
x^2-9 = x^2 -16x/3 +5/3
16x/3=32/3
x=2

 

[SoulSearcher]

ummm .. ive got another few questions which ii would like help with .. ><"

log(x^2 - 9) - log(x - 5) = log(x - 1/3)


log *base 2* (2^(2x) - 56) - x = 0


5^(2x) - 5^(x+1) = -4


ii still dont what to do first !! ><" ..

1) log{(x²-9)/(x-5)} = log(x- 1/3)
(x²-9)/(x-5) = (x- 1/3)
x²-9 = (x-5)(x-1/3)
x²-9 = x² - 16x/3 + 5/3
16x/3 = 32/3
16x = 32
x = 2

2)log₂(2^2x-56) - x = 0
2^2x - 2^x - 56 = 0
let u = 2^x
u² - u - 56 = 0
(u+7)(u-8) = 0
Therefore u = -7 or u = 8
Therefore 2^x = -7 or 2^x = 8 = 2³
But you cannot have 2^x = -7, therefore
2^x = 8 = 2³
Therefore x = 3

3) 5^2x - 5^x+1 = -4
5^2x - 5^x+1 + 4 = 0
5^2x - 5*5^x + 4 = 0
Let u = 5^x
u² - 5u + 4
(u-1)(u-4) = 0
Therfore u = 1 or u = 4
Therefore 5^x = 1 or 5^x = 4
Therefore x = 0 or x = log₅4

 

[ice ken]

1) log{(x²-9)/(x-5)} = log(x- 1/3)
(x²-9)/(x-5) = (x- 1/3)
x²-9 = (x-5)(x-1/3)
x²-9 = x² - 16x/3 + 5/3
16x/3 = 32/3
16x = 32
x = 2

2)log₂(2^2x-56) - x = 0
2^2x - 2^x - 56 = 0
let u = 2^x
u² - u - 56 = 0
(u+7)(u-8) = 0
Therefore u = -7 or u = 8
Therefore 2^x = -7 or 2^x = 8 = 2³
But you cannot have 2^x = -7, therefore
2^x = 8 = 2³
Therefore x = 3

3) 5^2x - 5^x+1 = -4
5^2x - 5^x+1 + 4 = 0
5^2x - 5*5^x + 4 = 0
Let u = 5^x
u² - 5u + 4
(u-1)(u-4) = 0
Therfore u = 1 or u = 4
Therefore 5^x = 1 or 5^x = 4
Therefore x = 0 or x = log₅4

too smart couldnt do 2 n 3 sooooooooooo hard

 

[SoulSearcher]

Those weren't as hard as you make it out to be.

 

[ice ken]

Those weren't as hard as you make it out to be.

they were alrite. even this yr 12 maths is easier. i didnt even know where to start for question 2 n 3. fogot all log rulz

 

[z600]

They are out of North Shore papers arent they ^^

log x^2-9-logx-5=log(x-1/3) cancel out the logs

(x^2-9)/(x-5)=3x-1/3 cross multiply

3x^2-27=(3x-1)(x-5)

3x^2-16x+6=3x^2-27 cancel out 3x^2

14x+6=27

x=4.6 sub 4.6 in, because 4.6-5 is equal to a negative number, and since u cant log negative number there are no solutions. (that was a nasty question)

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log (base 2) (2^2x-56)-x=0

2^x=2^2x-56 change back to index laws

let a be a=2^x a substitution

a=a^2-56

a^2-x-56=0

(a-8)(a+7)=0

x^2=8 x^2=-7 put x back in, remeber a=2^x

x=squareroot 8 -7 is not the anwer, u can sqaureroot negative numbers

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5^2x-5^x+1=-4

5^2x-5^x * 5=-4

let a be 5^x

a^2-5a+4=0

a^2-5a+4=0

(a-4)(a+1)

a= 4

x=squaroot 4 sub back

x=2 other answer is negative, cant be squared



out of all i reckon 4b was the hardest question^^

 

[z600]

Soul Searcher, mind having a go at this one, half the year 12 students cant do this one and yes i go to a comprehensive high school


3^2x*5^x=7^3x-1

 

[sle3pe3bumz]

LOLS .. damnn .. thanks guys .. ahah .. and yeah they were from the north shore hw .. ahah .. ii found 4b alright .. ahah .. ><" .. stupid north shore !! =D