Question 1209558
Here's how to solve these problems:

**a) 7^(sin²x) - 7^(cos²x) = 8**

1.  Use the identity cos²x = 1 - sin²x:

7^(sin²x) - 7^(1-sin²x) = 8

2.  Rewrite the second term:

7^(sin²x) - 7 / 7^(sin²x) = 8

3.  Let y = 7^(sin²x):

y - 7/y = 8

4.  Multiply by y:

y² - 7 = 8y

5.  Rearrange into a quadratic equation:

y² - 8y - 7 = 0

6.  Factor:

(y - 7)(y + 1) = 0

7.  Solve for y:

y = 7 or y = -1

Since 7^(sin²x) must be positive, y = -1 is not a valid solution.  So, y = 7.

8.  Substitute back:

7^(sin²x) = 7¹

sin²x = 1

9.  Solve for x:

sin x = ±1
x = (2n + 1)π/2, where n is an integer.

**b) [x/(x+1)]² + [x/(x-1)]² = m² + m**

1.  Simplify the left side:

x²/(x²+2x+1) + x²/(x²-2x+1) = m² + m

2. Find a common denominator and combine the fractions:

[x²(x²-2x+1) + x²(x²+2x+1)] / [(x²+2x+1)(x²-2x+1)] = m² + m

[x⁴-2x³+x² + x⁴+2x³+x²] / (x⁴-2x²+1) = m² + m

(2x⁴ + 2x²) / (x⁴ - 2x² + 1) = m² + m

2x²(x² + 1) / (x² - 1)² = m(m + 1)

This equation is a bit complex.  Without a specific value for 'm', it's difficult to simplify further to directly solve for x.  The best approach would be to substitute a given value for m and then try to solve for x.

**c) ab = ½, bc = ⅓, ac = 1/6**

1. Multiply the three equations together:

(ab)(bc)(ac) = (1/2)(1/3)(1/6)
a²b²c² = 1/36

2. Take the square root of both sides:

abc = ±1/6

3. Divide (abc = ±1/6) by each of the original equations to get the reciprocals:

1/c = (abc)/(ab) = (±1/6)/(1/2) = ±1/3, so c² = 9
1/a = (abc)/(bc) = (±1/6)/(1/3) = ±1/2, so a² = 4
1/b = (abc)/(ac) = (±1/6)/(1/6) = ±1, so b² = 1

4. Calculate the sum of the reciprocals squared:

(1/a²) + (1/b²) + (1/c²) = 1/4 + 1 + 9 = 10.25 = 41/4

**d) 2^x - 3^y = 5 and 2^(x+2) + 3^(y+2) = 59**

1. Rewrite the second equation:

4 * 2^x + 9 * 3^y = 59

2. Let u = 2^x and v = 3^y. The system becomes:

u - v = 5
4u + 9v = 59

3. Solve for u and v: From the first equation, u = v + 5. Substituting this into the second equation:

4(v + 5) + 9v = 59
13v + 20 = 59
13v = 39
v = 3

Then, u = v + 5 = 3 + 5 = 8.

4. Substitute back to find x and y:

2^x = 8 = 2³  => x = 3
3^y = 3 = 3¹  => y = 1

5. Find xy:

xy = 3 * 1 = 3

**e) 9^(4^m) = 4^(9^m)**

1. Take the logarithm of both sides (any base, but let's use the natural log):

ln(9^(4^m)) = ln(4^(9^m))

2. Use the logarithm power rule:

4^m * ln(9) = 9^m * ln(4)

3. Rewrite ln(9) as 2ln(3) and ln(4) as 2ln(2):

4^m * 2ln(3) = 9^m * 2ln(2)

4. Simplify:

4^m * ln(3) = 9^m * ln(2)

5. Rearrange:

4^m / 9^m = ln(2) / ln(3)

(2²/3²)^m = ln(2) / ln(3)

(2/3)^(2m) = log₃2

2m * ln(2/3) = ln(log₃2)

m = ln(log₃2) / (2 * ln(2/3))

**f) (7^(log₈x)) * (x^(log₉x)) = 3969**

1. Take the logarithm base 8 of both sides:

log₈[(7^(log₈x)) * (x^(log₉x))] = log₈3969

2. Use logarithm properties:

log₈(7^(log₈x)) + log₈(x^(log₉x)) = log₈3969

(log₈x)(log₈7) + (log₉x)(log₈x) = log₈3969

3. Let y = log₈x:

y * log₈7 + (log₉x) * y = log₈3969

y(log₈7 + log₉x) = log₈3969

We also know that 3969 = 63² = 9*7*9*7 = 3⁴ * 7²

log₈3969 = log₈(3⁴ * 7²) = 4log₈3 + 2log₈7

y(log₈7 + log₉x) = 4log₈3 + 2log₈7

This equation is still quite complex. It's likely there's a clever substitution or manipulation I'm missing to simplify it further and solve directly for x. Numerical methods or approximations might be necessary.