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Urgent help please ! Logs / intersections point

Need some help please :smile: I would be very appreciative of any help. Thanks in advance.


The curves y = 3^(x+2) and y = 5^(x-1) meet at a point Find the point of intersection of where the two curves meet.

Edit: I'm sooo sorry I made a typo, it should have been this instead 😁😩
(edited 8 years ago)
Reply 1
Avatar for TeeEm
TeeEm
19
Original post by Aty100
Need some help please :smile: I would be very appreciative of any help. Thanks in advance.


The curves y = 3^(x+1) and y = 5^(x-1) meet at a point Find the point of intersection of where the two curves meet.


solve their equations simulatenously
Reply 2
Avatar for Aty100
Aty100
OP
15
Original post by TeeEm
solve their equations simulatenously


With logs?

What is did was equated them and then did log base of 3 and solved
I got X to be -2.738
I don't think that works :/
(edited 8 years ago)
Reply 3
Avatar for Notnek
Notnek
21
Original post by Aty100
With logs?

What is did was equated them and then did log base of 3 and solved
I got X to be -2.738
I don't think that works :/

That's not right. Can you post all your working?
Reply 4
Avatar for Aty100
Aty100
OP
15
Original post by notnek
That's not right. Can you post all your working?


Thanks for helping :smile:
image.png250.9KB
Original post by Aty100
Need some help please :smile: I would be very appreciative of any help. Thanks in advance.


The curves y = 3^(x+1) and y = 5^(x-1) meet at a point Find the point of intersection of where the two curves meet.


(x+1)log⁡33=(x−1)log⁡35 (x+1) \log_3{3} = (x-1) \log_3{5}

x(1−log⁡35)=−1−log⁡35 x(1 - \log_3{5}) = -1 - \log_3{5}

x=−1+log⁡351−log⁡35 x = - \dfrac{1 + \log_3{5}}{1 - \log_3{5}}

Remember that you can still use the log rules even if the bases are weird like this one. Maybe it would be neater to have done it with natural logs.
Original post by Aty100
Thanks for helping :smile:


I'm afraid that is not the question that you posted originally.
Reply 7
Avatar for Notnek
Notnek
21
Original post by Aty100
Thanks for helping :smile:

That's a bit different to the question you posted - which one is correct?

There's a mistake in your working:

x+2−xlog⁡35=x(2−log35)x + 2 - x\log_3 5 = x(2-log_3 5)

If you expand the right-hand-side you'll see that this is wrong.

Instead, move the 2 to the other side and then take out a factor of x.
Reply 8
Avatar for Aty100
Aty100
OP
15
Original post by Louisb19
(x+1)log⁡33=(x−1)log⁡35 (x+1) \log_3{3} = (x-1) \log_3{5}

x(1−log⁡35)=−1−log⁡35 x(1 - \log_3{5}) = -1 - \log_3{5}

x=−1+log⁡351−log⁡35 x = - \dfrac{1 + \log_3{5}}{1 - \log_3{5}}

Remember that you can still use the log rules even if the bases are weird like this one. Maybe it would be neater to have done it with natural logs.


OMG I TYPED MY QUESTION OUT WRONG!!
I'm Sooo sorry.

It's supposed to be 3^(X+2)
Reply 9
Avatar for Aty100
Aty100
OP
15
Original post by notnek
That's a bit different to the question you posted - which one is correct?

There's a mistake in your working:

x+2−xlog⁡35=x(2−log35)x + 2 - x\log_3 5 = x(2-log_3 5)

If you expand the right-hand-side you'll see that this is wrong.

Instead, move the 2 to the other side and then take out a factor of x.


Sorry it should have been 3^x+2
Reply 10
Avatar for Notnek
Notnek
21
Original post by Aty100
Sorry it should have been 3^x+2

Okay. My last post explains where you went wrong.
Reply 11
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Aty100
OP
15
Original post by notnek
Okay. My last post explains where you went wrong.



I don't understand why it is x(2-log3 5)
Where did the 2 come from ?
Reply 12
Avatar for Notnek
Notnek
21
Original post by Aty100
I don't understand why it is x(2-log3 5)
Where did the 2 come from ?

You wrote x(2-log3 5) in your working.

I'm telling you that it's incorrect.
Reply 13
Avatar for Aty100
Aty100
OP
15
Original post by notnek
You wrote x(2-log3 5) in your working.

I'm telling you that it's incorrect.


Oh sorry.

So

X+2 = x(log3 5) - 1 ( log3 5)

X - Xlog3 5 = -1log3 5 -2

X(1-log3 5) = ( -1log3 5 ) - 2

X = ( -1log3 5 ) - 2 / (1-log3 5)

??
Reply 14
Avatar for KaylaB
KaylaB
20
Original post by Aty100
Need some help please :smile: I would be very appreciative of any help. Thanks in advance.


The curves y = 3^(x+2) and y = 5^(x-1) meet at a point Find the point of intersection of where the two curves meet.

Edit: I'm sooo sorry I made a typo, it should have been this instead 😁😩


As suggested above it would be easier to take logs base 10, rather than stressing and getting confused about log base 3
Reply 15
Avatar for Notnek
Notnek
21
Original post by Aty100
Oh sorry.

So

X+2 = x(log3 5) - 1 ( log3 5)

X - Xlog3 5 = -1log3 5 -2

X(1-log3 5) = ( -1log3 5 ) - 2

X = ( -1log3 5 ) - 2 / (1-log3 5)

??

That's correct.
Reply 16
Avatar for Aty100
Aty100
OP
15
Original post by notnek
That's correct.


I got X= 7.4 but that doesn't work when I sub it back into the y equations 😩
Reply 17
Avatar for KaylaB
KaylaB
20
Original post by Aty100
Oh sorry.

So

X+2 = x(log3 5) - 1 ( log3 5)

X - Xlog3 5 = -1log3 5 -2

X(1-log3 5) = ( -1log3 5 ) - 2

X = ( -1log3 5 ) - 2 / (1-log3 5)

??



For an alternative method
Reply 18
Avatar for Notnek
Notnek
21
Original post by Aty100
I got X= 7.4 but that doesn't work when I sub it back into the y equations 😩

It's correct and it satisfies the equations if you use the exact value of x i.e. not the rounded value 7.4. (actually 7.5 is correct for 1 d.p.)
Reply 19
Avatar for Aty100
Aty100
OP
15
Original post by notnek
It's correct and it satisfies the equations if you use the exact value of x i.e. not the rounded value 7.4. (actually 7.5 is correct for 1 d.p.)


Thank SOO much !

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