Inverse of Exponential Equations

[Murdoc88]

Alright, I'm just going through my older tests in preparation for my upcoming exams and I have come across a question regarding exponents and inverses.

Homework Statement

The original equation is:

h(x) = 3(4^{(x-1)})+2

and I must find the inverse in order to proceed with the rest of the question.

Homework Equations

Now I know the basic stuff of Logs, Natural Logarithms and Exponents such as the basic rules:

Change of Base
Product
Quotient
Power

and I can do the other questions on the test but it is this one question with the additional constants that is throwing my mathematical abilities out of the window

If someone could just give me some guidance on how to approach this problem I would be eternally grateful.

- Murdoc

 

[Murdoc88]

Also inside the brackets it is suppose to be 4^(x-1). I'm still new with the Latex coding

 

[cristo]

So you have some equation of the form y=3(4^{(x-1)})+2. To find this inverse of this function, you need to make x the subject of the formula. So, firstly, subtract 2 and divide by 3 to give \frac{y-2}{3}=4^{(x-1)}. Now, you can take the natural log of both sides, and use rules of logarithms to simplify. Alternatively, you could use logarithm with base four. Can you go on from here?

 

[Murdoc88]

I'm still confused as I've never done a question to this extent before, so it's still a relatively new concept to understand.

If I take the Log base 4 route I end up with this:

(x-1) = Log(4)((y-2)/3)
X-1 = -3Log(4)(Y-2)

Is this correct or am I missing something? and if that's correct, where would I go from here?

X=-3Log(4)(Y-2)+1?

 

[Vagrant]

log a/b = log a - log b

 

[cristo]

I'm still confused as I've never done a question to this extent before, so it's still a relatively new concept to understand.

If I take the Log base 4 route I end up with this:

(x-1) = Log(4)((y-2)/3)
X-1 = -3Log(4)(Y-2)

Is this correct or am I missing something? and if that's correct, where would I go from here?

X=-3Log(4)(Y-2)+1?

To express shramana's point more succintly; this line is correct:
x-1 = \log_4(\frac{y-2}{3})
However, your next line is incorrect. You could simplify this using the logarithm of quotients rule then add 1 to both sides, or could just leave it as log((y-2)/3) and add 1 to both sides. It's up to you really.