Question Help Needed... (1 Viewer)

[shaon0]

I need help with this question....
--For the first question i tried:
(3^x+1 + 3^x-1)/3^x
=(x+1.ln(3) + x-1 ln(3))/ x.ln(3)
=ln(3)(x+1+x-1) / ln(3).x
=2x/x
=2

--For the second question i tried:
625^x+y = 1/5
x+y.ln(625) = ln(1/5)
x+y = -1/4
Thus, y= -(1+4x) /4 ....(1A)

32.2^x = 4^y.....(2)
64^x = 4^y
x. ln(64) = y.ln (4)
x = y/3
Thus, y = 3x...(2A)

Sub. (1A) into (2A)
3x = - (1+4x)/4
12x = -1-4x
16x = -1
Thus, x= -1/16

Sub. x value into (2)
32.2^-1/16 = 4^y
Thus, y= (32.2^-1/16) / 4


Any help would be appreciated....

 

[lyounamu]

I need help with this question.... For the first question i tried:
(3^x+1 + 3^x-1)/3^x
=(x+1.ln(3) + x-1 ln(3))/ x.ln(3)
=ln(3)(x+1+x-1) / ln(3).x
=2x/x
=2

Any help would be appreciated....

a)
3^(x+1) + 3^(x-1)/3^x = (3 . 3^x + 3^x . 1/3)/3^x
= (3^x)(3+1/3)/(3^x)
= 3 + 1/3 = 10/3

c) 125^x . 5^y = 1/5
(5^3)^x . 5^y = 5^(-1)
5^(3x+y) = 5^(-1)

Therefore, 3x+y = -1 ...(1)

Now, 2^x = 4^y/32
2^x = ((2^2)^y)/2^5
2^x = 2^(2y-5)
Therefore, x=2y-5

Substitue 2y-5 for the value of x into the equation (1)
3x+y = -1
3(2y-5)+y = -1
6y-15 + y = -1
7y = 14
y =2
Substitue 2 for the value of y into the equation (1) to find the value of x.
Therefore, x = -1

 

[minijumbuk]

I'm pretty sure this is 2 unit work, but anyway...

1. 3.3^x + 3^-1.3^x
3^x
= 3^x(3+ 1/3)
3^x
=3+1/3
=10/3

2.
5^3x.5^y = 5^-1
5^3x+y = 5^-1
3x+y = -1 ------------1

2⁵.2^x = 2^2y
2^x+5 = 2^2y
x+5 = 2y
x = 2y - 5 -------------2

Sub 2 into 1
3(2y-5) + y = -1
6y - 15 + y + 1= 0
7y=14
y=2
Since x+5=2y,
x= 2y - 5
x= 2(2)-5
x= -1

Therefore x=-1, y = 2.

Damn, lyounamu, you beat me this time xD

 

[lyounamu]

I'm pretty sure this is 2 unit work, but anyway...

1. 3.3^x + 3^-1.3^x
3^x
= 3^x(3+ 1/3)
3^x
=3+1/3
=10/3

2.
5^3x.5^y = 5^-1
5^3x+y = 5^-1
3x+y = -1 ------------1

2⁵.2^x = 2^2y
2^x+5 = 2^2y
x+5 = 2y
x = 2y - 5 -------------2

Sub 2 into 1
3(2y-5) + y = -1
6y - 15 + y + 1= 0
7y=14
y=2
Since x+5=2y,
x= 2y - 5
x= 2(2)-5
x= -1

Therefore x=-1, y = 2.

Damn, lyounamu, you beat me this time xD

LOL. What an incredible coincidence.

I only beat you by few seconds. And your working-out takes longer (looking at all the indexes and etc) so you did better than I did.

 

[shaon0]

Thanks guys i knew the answer but i was wondering why my way wasn't correct for the first question....
Could someone explain?

 

[lyounamu]

Thanks guys i knew the answer but i was wondering why my way wasn't correct....
Could someone explain?

a) If you are using log3, you have to apply the log 3 to the fraction as a whole fraction, not as individuals.

b) 5^y . 125^x =/= 625^xy. It's against the Index Law. Index Law states that: a^b . a^c = a^(b+c) but a^b . c^d =/= ac^bd.

You need to have the same base for your working-out to work.

 

[shaon0]

a) If you are using log3, you have to apply the log 3 to the fraction as a whole fraction, not as individuals.

b) 5^y . 125^x =/= 625^xy. It's against the Index Law. Index Law states that: a^b . a^c = a^(b+c) but a^b . c^d =/= ac^bd.

You need to have the same base for your working-out to work.

Thanks, so i can't apply them individual. I thought my method was pretty elegant and quick so i thought i'd be right..
This was from a 3unit paper by the way.

 

[lyounamu]

Thanks, so i can't apply them individual. I thought my method was pretty elegant and quick so i thought i'd be right..
This was from a 3unit paper by the way.

Good practice question, that is.

 

[shaon0]

Good practice question, that is.

Yea i know...My school's tests are notorious for questions involving logs and other random topics which we haven't covered properly.

 

[minijumbuk]

Which school do you go to, may I ask?

 

[tommykins]

回复: Re: Question Help Needed...

Yea i know...My school's tests are notorious for questions involving logs and other random topics which we haven't covered properly.

this isn't really logs, just index laws

 

[shaon0]

Which school do you go to, may I ask?

Patrician Brothers Blacktown.