Expert: Sherman D. - 1/29/2008

Question
solve the equations for x, a real number.
(sqrt 2 + 1)^x - (sqrt 2 - 1) ^x =14
and
6^2x+4=(2^x+8) * (3^3x)
for the second one I know you can combine the exponents and then they would both have the same base but when you solve it and put the solution back in (-2) it doesn't work.

Answer
(sqrt(2) + 1)^x - (sqrt(2) - 1)^x = 14

ANS : 3

i found the answer out by graphing at http://www.calculator.com/calcs/GCalc.html

i also asked Yahoo Answers to see if they could find the answer some way, and here is what they came up with.

The cannot be solved algebraically , therefore you will have to solve it numerically .
(sqrt(2) + 1 )^x - (sqrt(2) - 1)^x = 14
(sqrt(2) + 1)^x = 14 + (sqrt(2) - 1)^x
ln both sides .
xln(sqrt(2) + 1) = ln(14 + (sqrt(2) - 1)^x )
x = ln(14 + (sqrt(2) - 1)^x ) / ln(sqrt(2) + 1)

Now this series is a convergence series .
X(n + 1) = ln(14 + (sqrt(2) - 1)^xn ) / ln(sqrt(2) + 1)

Start with x0 =1
x1 = 3.027336889
x2 = 2.9999863573
x3 = 3.00000000689
x4 = 2.9999999997
x5 = 3
x6 = 3
Now you stop
You see x5 and x6 are both = 3

There x = 3

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

6^(2x + 4) = 2^(x + 8) * 3^(3x)
(3 * 2)^(2x + 4) = 2^(x + 8) * 3^(3x)
2^(2x + 4) * 3^(2x + 4) = 2^(x + 8) * 3^(3x)

for this one to work

2x + 4 = x + 8
and also
2x + 4 = 3x

so this now becomes

x + 8 = 3x
-2x = -8
x = 4

ANS : x = 4