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 +

Algebra ->  Exponential-and-logarithmic-functions -> 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 +       Log On


   

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) About Me  (Show Source):
You can put this solution on YOUR website!
16%5E%28x%29+-+4%5E%28x-1%29+-+3+=+0

%284%5E2%29%5Ex+-+4%5E%28x-1%29+-+3+=+0

4%5E%282x%29+-+4%5E%28x-1%29+-+3+=+0

%284%5Ex%29%5E2+-+4%5E%28x-1%29+-+3+=+0

%284%5Ex%29%5E2+-+%284%5Ex%29%2A%284%5E%28-1%29%29+-+3+=+0

%284%5Ex%29%5E2+-+%284%5Ex%29%2A%281%2F4%29+-+3+=+0

%284%5Ex%29%5E2+-+%281%2F4%29%284%5Ex%29+-+3+=+0

z%5E2+-+%281%2F4%29z+-+3+=+0 Let z+=+4%5Ex

4z%5E2+-+z+-+12+=+0 Multiply both sides by the LCD 4

Use the quadratic formula to get z+=+%281%2Bsqrt%28193%29%29%2F8 or z+=+%281-sqrt%28193%29%29%2F8

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.

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The solutions in terms of z are z+=+%281%2Bsqrt%28193%29%29%2F8 or z+=+%281-sqrt%28193%29%29%2F8

We ultimately want solutions in terms of x.

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

z+=+4%5Ex

%281%2Bsqrt%28193%29%29%2F8+=+4%5Ex

4%5Ex+=+%281%2Bsqrt%28193%29%29%2F8

x+=+log%284%2C%28%281%2Bsqrt%28193%29%29%2F8%29%29

So the first solution is x+=+log%284%2C%28%281%2Bsqrt%28193%29%29%2F8%29%29

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

z+=+4%5Ex

%281-sqrt%28193%29%29%2F8+=+4%5Ex

4%5Ex+=+%281-sqrt%28193%29%29%2F8

x+=+log%284%2C%28%281-sqrt%28193%29%29%2F8%29%29

So the second solution is x+=+log%284%2C%28%281-sqrt%28193%29%29%2F8%29%29

The exact solutions for x are: x+=+log%284%2C%28%281%2Bsqrt%28193%29%29%2F8%29%29 or x+=+log%284%2C%28%281-sqrt%28193%29%29%2F8%29%29

Use a calculator to get these approximate solutions x+=+0.44825431399819 or x+=+UND

the UND means "undefined" so it turns out that the second solution x+=+log%284%2C%28%281-sqrt%28193%29%29%2F8%29%29 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 x+=+log%284%2C%28%281%2Bsqrt%28193%29%29%2F8%29%29

The only approximate solution is x+=+0.44825431399819


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