SOLUTION: Solve 16^(x) - 4^(x-1) - 3 = 0 By looking at the problem it seems to me this is an exponential equation of the quadratic flavor which can somehow can be rewritten in the ax^2 +

Question 913131: Solve 16^(x) - 4^(x-1) - 3 = 0
By looking at the problem it seems to me this is an exponential equation of the quadratic flavor which can somehow can be rewritten in the ax^2 + bX + c = 0 fashion. But I don't get much farther than turning the 16^x into a (4^x)^2. This has stumped me. I've tried different things and at the end I have to resort to using the quadratic formula to factor this bugger, but it never results close to the actual answer of logbase4(-2 + √7)
Any help is much appreciated!
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!

Let

Multiply both sides by the LCD 4

Use the quadratic formula to get or

The steps for the quadratic formula are lengthy and I feel they would clutter up this solution page. Let me know if you need to see the steps to getting those solutions for z.

-----------------------------------------------------------------------------------

The solutions in terms of z are or

We ultimately want solutions in terms of x.

Plug in the first solution for z to get the first solution for x.

So the first solution is

----------------------------------------

Plug in the second solution for z to get the second solution for x.

So the second solution is

The exact solutions for x are: or

Use a calculator to get these approximate solutions or

the UND means "undefined" so it turns out that the second solution isn't really a solution at all. The argument isn't in the domain of log(x)

So let's revise our answer. The only exact solution is

The only approximate solution is

Let me know if you need more help or if you need me to explain a step in more detail.
Feel free to email me at jim_thompson5910@hotmail.com
or you can visit my website here: http://www.freewebs.com/jimthompson5910/home.html

Thanks,

Jim

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