Question

1. Solve for x:9^x*2^2x=6

Answer 1

This can be read in different ways:

1. (9^x)*(2^2)x=6: (9^x)(4x)=6, (9^x)x=3/2=1.5, x=0.5 or 1/2;
2. (9^x)*(2^(2x))=6: x=0.5 or 1/2
3. (9^(x*2^2))*x=6: (9^(4x))x=6, x=0.330007 approx.

There may be other interpretations but I'm assuming that the simplest answer is the one required.

The results I got were partly trial and error, and the best answer, which fits two interpretations, is x=0.5. The clue to working it out is to realise that 9 is a perfect square so that 9^(1/2)=3 and 3 goes into 6 twice. The number 2 can be obtained by 2^2x=4*(1/2) or by 2^(2x)=2^1=2 when x=1/2. The third answer is totally trial and error and is probably not applicable.

Answer 2

Assuming that the function is: 9^(x*2^(2x)) = 6,

There are two ways of doing this.

1st Way

Simply graph the function above and read off the graph.

Plot the curve y = 9^(x*2^(2x)) and the straight line y = 6 and see where they intersect.

You will find that there are two points of intersection, at x = -0.442 and x = 0.442.

2nd Way

The function is,

9^(x*2^(2x)) = 6

Taking  logs of both sides,

(x*2^(2x)) = ln(6)/ln(9)

Again, taking  logs of both sides,

ln(x) + 2x.ln(2) + ln(ln(9)/ln(6)) = 0

Rewrite the function as f(x) = ln(x) + 2x.ln(2) + ln(ln(9)/ln(6))

Now use the Newton-Raphson method to find the solution for f(x) = 0

Using f’(x) = 1/x + 2.ln(2),

The iterative equation is: x_(k+1) = x_k – f(x_k) / f’(x_k)

Taking x1 = 0.5, the table of results is,

n       x_n                      f(x_n)                     f’(x_n)                   x_(n+1)

1       0.5                       0.2039969278      3.386294361     0.4397580641

2      0.4397580641  -0.0078995052      3.660271990     0.4419162385

3      0.4419162385  -0.0000120031      3.649166627     0.4419195278

4       0.4419195278  0.00                        3.649149784     0.4419195278

5       0.4419195278  0.00                        3.649149784     0.4419195278

We achieve a result to 10 dp after 5 iterations, which result conforms closely to the graphical result.

We can now write down the result(s) as: x = +/- 0.4419195278