Solving Exponential Equations Using Logarithms

[enkrypt0r]

Solve for x:

(4^x)-(6*2^x) = -8

How would I go about starting this? At first I tried taking the ln of each side to get the variable out from the exponent, but I get pretty confused as to what the result would be. Would it be anything like the following?

2x + ln(4) - (ln(6) * ln(2)) = ln(-8)

Am I close, or am I completely missing it?

I just plain don't know where to start. I've been working on the sheet for a few hours, trying something different each time, and nothing seems to end up right. I would really appreciate a push in the right direction.

I know the properties I just don't really know what order to use them in.

There's a few other questions that I'm having trouble with, such as:

6e^(2x) - 16e^x = 6

log(4x-1) = log(x+1) + log(2)Like I said before, I know the properties, so you don't need to tell me those, I would just like to know when and how to use them properly. I don't even necessarily need the answers, just sort of a walkthrough.

Your help is appreciated, this is the only thing in precalc that I can't seem to grasp.

 

[Mark44]

Taking the log of each side is no help. There is no property of logs that works with sums or differences. For your problem, you can take the log of the left side to get
ln(4^x - 6* 2^x), but there's nothing more you can do.
It's even worse on the right side, since ln(-8) is undefined. That should have been a clue.

The equation you have is quadratic in form. 4^x = (2^x)*(2^x) = (2^x)^2. I think you will find that the equation can be factored, and you can go from there.

 

[HallsofIvy]

If you let y= 2^x, what is 4^x? You should be able to get, and solve, a polynomial equation in y, then find x.