Exponential Equations: Urgent Help

Set your parentheses right about exponents before expecting help

 

Firstly, it would help if you put brackets around the exponents. For example, is question 1. 9^2x=27^x-1 or 9^2x=27^x-1? I presume it is the former, but I disagree with your answer. Show your work; what base would you choose to take on each side? What equation does this yield?

 

Make your equation EXPLICIT at this point!
Otherwise, we can't help you along..

 

No, it is not right, not your notation, and not your expression of 9, nor your contraction into an exponent equation.
We have:
[tex]9^{2x}=27^{(x-1)}, 9=3^{2},27=3^{3}\to(3^{2})^{2x}=(3^{3})^{x-1}\to3^{2*(2x)}=3^{3*(x-1)}[/tex]
yielding the exponent equation:
[tex]2*(2x)=3*(x-1)[/tex]

 

No. It is easy to verify that [itex]9^{(2x)} \ne 3^{(4(2x)}[/itex]

 

Only if you believe -4=-6.

Do you understand what's going on here? We're saying that 9 can be represented by 3^2. Therefore, 9^x=(3^2)^x=3^(2x). Now the trick to solving these is to get them to the same base to get an 'exponent' equation.

 

You are guessing (you original post) and being sloppy (your last post). What is [itex]3*(x-1)[/tex]?

 

You expand 3(x-1) as 3x-1. This is not correct. It should be 3(x-1)=3x-3

 

Do you understand how to multiply out a parenthesis?

 

Okay, you're doing quite a few things wrong. Please don't be so liberal with your equality symbols. Break it up. Go the long way. Don't do shortcuts, this is where you're getting messed up.

9^(2x)=27^(x-1) iff (3^2)^(2x)=(3^3)^(x-1) iff 3^(4x)=3^(3x-3) iff 4x=3x-3 iff x=-3.

The key thing to realize here is that (a^n)^m=a^(nm). (Where these formulas make sense...of course :| )

 

Don't thank me, thank arildno. He already told you precisely what I did. Except I gave you a final answer...which I'm kind of regretting.