We are asked to solve four different math problems: (a) Find the value of $w$ such that $\log(w+5) + \log(w-5) = 4\log(2) + 2\log(3)$. (b) Solve for $x$ in the equation $(\frac{1}{3})^{\frac{x^2-2x}{16-2x^2}} = \sqrt[4x]{9}$. (c) Simplify the expression $[5a^5b^2 \times 3(ab^3)^2] \div (15a^2b^8)$. (d) If $a$ and $b$ are whole numbers such that $a^b = 121$, evaluate $(a-1)^{b+1}$.

1. Problem Description

We are asked to solve four different math problems:

(a) Find the value of $w$ such that $\log(w+5) + \log(w-5) = 4\log(2) + 2\log(3)$.

(b) Solve for $x$ in the equation $(\frac{1}{3})^{\frac{x^2-2x}{16-2x^2}} = \sqrt[4x]{9}$.

(c) Simplify the expression $[5a^5b^2 \times 3(ab^3)^2] \div (15a^2b^8)$.

(d) If $a$ and $b$ are whole numbers such that $a^b = 121$, evaluate $(a-1)^{b+1}$.

2. Solution Steps

(a)

We have the equation $\log(w+5) + \log(w-5) = 4\log(2) + 2\log(3)$.

Using the logarithm properties $\log(A) + \log(B) = \log(AB)$ and $n\log(A) = \log(A^n)$, we get

$\log((w+5)(w-5)) = \log(2^4) + \log(3^2)$.

$\log(w^2-25) = \log(16) + \log(9)$.

$\log(w^2-25) = \log(16 \times 9)$.

$\log(w^2-25) = \log(144)$.

Since the logarithms are equal, we have $w^2-25 = 144$.

$w^2 = 144+25 = 169$.

$w = \pm \sqrt{169} = \pm 13$.

Since we need $w+5 > 0$ and $w-5 > 0$, we require $w > 5$.

Therefore, $w = 13$.

(b)

The equation is $(\frac{1}{3})^{\frac{x^2-2x}{16-2x^2}} = \sqrt[4x]{9}$.

$(\frac{1}{3})^{\frac{x^2-2x}{16-2x^2}} = 9^{\frac{1}{4x}}$.

$(3^{-1})^{\frac{x^2-2x}{16-2x^2}} = (3^2)^{\frac{1}{4x}}$.

$3^{\frac{-x^2+2x}{16-2x^2}} = 3^{\frac{2}{4x}}$.

$3^{\frac{-x^2+2x}{16-2x^2}} = 3^{\frac{1}{2x}}$.

Therefore, $\frac{-x^2+2x}{16-2x^2} = \frac{1}{2x}$.

$\frac{-x(x-2)}{2(8-x^2)} = \frac{1}{2x}$.

$\frac{-x(x-2)}{2(2\sqrt{2}-x)(2\sqrt{2}+x)} = \frac{1}{2x}$.

$-x(x-2)(2x) = 16-2x^2$.

$-2x^2(x-2) = 16-2x^2$.

$-2x^3+4x^2 = 16-2x^2$.

$-2x^3+6x^2-16=0$.

$x^3-3x^2+8=0$.

Let $f(x) = x^3 - 3x^2 + 8$. We observe that $f(-1) = (-1)^3 - 3(-1)^2 + 8 = -1 - 3 + 8 = 4$.

$f(-2) = (-2)^3 - 3(-2)^2 + 8 = -8 - 12 + 8 = -12$.

$f(0) = 8$.

We can try $x=-1$, then $(\frac{1}{3})^{\frac{1+2}{16-2}} = (\frac{1}{3})^{\frac{3}{14}} = \sqrt[4(-1)]{9} = 9^{-\frac{1}{4}}$.

$(\frac{1}{3})^{\frac{3}{14}} = (\frac{1}{9})^{\frac{1}{4}}$

$3^{-\frac{3}{14}} = 3^{-\frac{2}{4}} = 3^{-\frac{1}{2}}$

This is not true.

However, if we try $x = -2$:

$(\frac{1}{3})^{\frac{4+4}{16-8}} = (\frac{1}{3})^{\frac{8}{8}} = \frac{1}{3}$.

$\sqrt[4(-2)]{9} = \sqrt[-8]{9} = 9^{-\frac{1}{8}} = (3^2)^{-\frac{1}{8}} = 3^{-\frac{1}{4}}$.

Still not true.

Notice that if $x^2-2x=0$, then $x=0$ or $x=2$. If $x=2$, then $\sqrt[8]{9}$ which is not $1$. If $x=0$ the root is not defined.

$\frac{-x(x-2)}{2(8-x^2)} = \frac{1}{2x}$.

$-x^2(x-2) = 8-x^2$.

$-x^3 + 2x^2 = 8-x^2$.

$x^3 - 3x^2 + 8 = 0$.

Let $x=-1.64$, then $(-1.64)^3 - 3(-1.64)^2 + 8 = -4.41 - 8.07 + 8 = -4.48$.

(c)

$[5a^5b^2 \times 3(ab^3)^2] \div (15a^2b^8)$.

$[5a^5b^2 \times 3a^2b^6] \div (15a^2b^8)$.

$[15a^7b^8] \div (15a^2b^8)$.

$\frac{15a^7b^8}{15a^2b^8} = a^{7-2}b^{8-8} = a^5b^0 = a^5$.

(d)

We are given $a^b = 121$, and $a, b$ are whole numbers.

Since $121 = 11^2$, we have $a=11$ and $b=2$.

We want to evaluate $(a-1)^{b+1} = (11-1)^{2+1} = (10)^3 = 1000$.

3. Final Answer

(a) $w=13$

(b) No real solutions can be easily found. The cubic equation $x^3 - 3x^2 + 8 = 0$ has one real root, approximately $x = -1.64$.

(c) $a^5$

(d) $1000$