Expert: Scott A Wilson - 1/6/2011

Question
Sir, I have question related to quadratic equations.

1.) 8x^(3/2) - 8x^(-3/2) = 63

2.) 2^(2x+3) - 9.2^x + 1 = 0

3.) 2^(2x+3) + 2^(x+3) = 1 + 2^x  [no value of x gives 2^x = -1]

4.) 2^2x - 3.2^(x+2) + 32 = 0

Sir, I hope I have been clear enough. Please do answer.
                                     regards,
                                     Prakash Kumar

Answer
1) The problem can be rewritten as 8x^1.5 - 8/x^1.5 = 64.
If x^0.5 is 2, then x^1.5 is 8 and 1/x^1.5 is 1/8.
With this, 8*8 - 8/8 = 64 - 1 = 63, so that's the answer.
So since x^0.5 = 2, x = 4.

2) If we have 2^(2x+3) - 9.2^x + 1 = 0, then 9.2^x = 2^(2x+3) + 1.
Taking the ln() of both sides gives x*ln(9.2) = ln(2^(x+3) + 1),
so a recursive relation to use is x[n+1] = ln(2^(x[n]+3) + 1)/ln(9.2).

Using Excel, this gives
0
0.990096038
1.273769434
1.357584609
1.382511566
1.389938114
1.392151863
1.392811851
1.393008623
1.39306729
1.393084782
1.393089997
1.393091552
1.393092016
1.393092154
1.393092195
1.393092207
1.393092211
1.393092212
1.393092213
1.393092213, and that's the answer since it reapeats.

3) Take the equation as 2^(2x+3) = 1 + 2^x - 2^(x+3).
Take the ln() of both sides and get (2x+3)ln(2) = ln(1 + 2^x - 2^(x+3)).
Divide by ln(2) and get 2x+3 = ln(1 + 2^x - 2^(x+3))/ln(2).
Subtract 3, giving 2x = ln(1 + 2^x - 2^(x+3))/ln(2) - 3.
Divide both sides by 2, giving x = {ln(1 + 2^x - 2^(x+3))/ln(2) - 3}/2.
Recursively, that is x[n+1] = {ln(1 + 2^x[n] - 2^(x[n]+3))/ln(2) - 3}/2.

I would try that and write back on any more questions.
I'm not sure what the fadct that 2^x = -1 has no solutions has to do with this.

4) On this problem, put the 2^2x on one side of the equation and the other terms on the other.
Take the ln() of both side, then divide both sides by 2*ln(2).
This should give a recursive relation to find x.

I could assist you further, but it's getting to late.
In fact, its so late that it is really morning.  Maybe I should go to bed sometime ...