Solve 27^(7x-4) = (1/81)^(8x-7); sqrt(t-131)-sqrt(t+103)=56

[HopefulMii]

Hey, I'm not sure whether this goes here or Calculus..but here goes, please help me ^^; I can't figure this out and it's due for tuesday!

Question 1:

Solve for x

Question 2:

Solve for t; the only possible root is t=?

I don't even know where to start for these problems...so if you could also include your steps and how you got to those answers that'd be great ^^ Thanks!

Mii

 

[Deleted member 4993]

Re: Homework Help

Hey, I'm not sure whether this goes here or Calculus..but here goes, please help me ^^; I can't figure this out and it's due for tuesday!

Question 1:

Solve for x
Use logarithm - use log base 3

Question 2:

Solve for t; the only possible root is t=?
Square both sides - and square again - judiciously


I don't even know where to start for these problems...so if you could also include your steps and how you got to those answers that'd be great ^^ Thanks!

Mii

If you are still stuck - write back showing your work - indicating exactly where you are stuck so that we know where to begin to help you.

 

[Loren]

Re: Homework Help

Hint on the first one.
27 = 3[sup:2codtflt]3[/sup:2codtflt]
1/81 = 1/3[sup:2codtflt]4[/sup:2codtflt] = 3[sup:2codtflt]-4[/sup:2codtflt]

On the second one.
Get one radical on the left side of the equation and the other on the other side then square both sides. You will still have one radical left, so get it all by itself on one side of the equation and all aother terms on the other side. Simplify and then square both sides again. That should eliminate the radical signs. Solve for the unknown. Be sure to check all solutions into the original equation because in the squaring process you possibly created extraneous solutions.

 

[HopefulMii]

Re: Homework Help

^^; Thanks for the quick response

I tried what you said but I'm still confused on what to do, hehe...

I attached what I did so far...

Q1: [attachment=1:h499zlcj]Picture 001.jpg[/attachment:h499zlcj]

and

Q2[attachment=0:h499zlcj]Picture 002.jpg[/attachment:h499zlcj]

 

[Loren]

Re: Homework Help

On the first one, if a[sup:265n3bqr]x[/sup:265n3bqr] = a[sup:265n3bqr]y[/sup:265n3bqr], then x = y.

On the second one, when you square a binomial, you get a trinomial.
(a+b)[sup:265n3bqr]2[/sup:265n3bqr] = a[sup:265n3bqr]2[/sup:265n3bqr]+2ab+b[sup:265n3bqr]2[/sup:265n3bqr].

 

[HopefulMii]

Re: Homework Help

Oh my god! Yay I got them right =]

Thanks alot, just a quick question though...

ax = ay, then x = y.

is that like a rule or something...?

 

[Loren]

Re: Homework Help

It's a rule but more common sense than anything else. Think about it. If I say 5^a = 5^b, can you come up with values for a and b that are not the same and have the original statement still be true? The only problem is if a[sup:zmjcpg2j]x[/sup:zmjcpg2j] = a[sup:zmjcpg2j]y[/sup:zmjcpg2j], then x = y, a cannot be 0.

 

[HopefulMii]

Re: Homework Help

It's a rule but more common sense than anything else. Think about it. If I say 5^a = 5^b, can you come up with values for a and b that are not the same and have the original statement still be true? The only problem is if a[sup:2nd22cmx]x[/sup:2nd22cmx] = a[sup:2nd22cmx]y[/sup:2nd22cmx], then x = y, a cannot be 0.

that makes sense, hehe. Thanks alot :mrgreen: :mrgreen:

 

[Deleted member 4993]

Re: Homework Help

It's a rule but more common sense than anything else. Think about it. If I say 5^a = 5^b, can you come up with values for a and b that are not the same and have the original statement still be true? The only problem is if a[sup:2l1mztmf]x[/sup:2l1mztmf] = a[sup:2l1mztmf]y[/sup:2l1mztmf], then x = y, a cannot be 0 [ or 1].