$\displaystyle 4 (x +8) + 13 = 6 (x + 9) - x$

$\displaystyle 4 \cdot x +4 \cdot 8 + 13 = 6 \cdot x + 6 \cdot 9 - x$

$\displaystyle 4 x +32 + 13 = 6x + 54 - x$

$\displaystyle 4 x +45 = 5x + 54$

$\displaystyle 4 x +45 -4x -54 = 5x + 54 -4x -54$

$\displaystyle -9 = x$

By substituting $\displaystyle u = x^{2}$ and, subsequently, $\displaystyle u^{2} =\left ( x^{2} \right )^{2} = x^{4}$, this equation be rewritten as a quadratic equation, and solved as such:

$\displaystyle x^{4} - 10x^{2} + 9 = 0$

$\displaystyle u^{2} - 10u + 9 = 0$

We are looking to factor the quadratic expression as $\displaystyle (u+?)(u+?)$, replacing the two question marks with integers with a product of  9 and a sum of $\displaystyle -10$; these integers are $\displaystyle -9,-1$.

$\displaystyle (u-9)(u-1) = 0$

Substitute back:

$\displaystyle (x^{2}-9)(x^{2}-1) = 0$

These factors can themselves be factored as the difference of squares:

$\displaystyle (x+3)(x-3)(x+1)(x-1) = 0$

Set each factor to zero and solve:

$\displaystyle x-3 = 0 \Rightarrow x = 3$

$\displaystyle x-1 = 0 \Rightarrow x = 1$

$\displaystyle x+1 = 0 \Rightarrow x = - 1$

$\displaystyle x+3 = 0 \Rightarrow x = - 3$

The solution set is $\displaystyle \left \{ -3, -1, 1, 3\right \}$.

Substitute $\displaystyle u = \sqrt{x}$, and, subsequently, $\displaystyle u^{2} = x$, to rewrite this equation as quadratic, then solve by factoring.

$\displaystyle x - 7 \sqrt{x} + 12 = 0$

$\displaystyle u^{2} - 7 u + 12 = 0$

We can rewrite the quadratic expression as $\displaystyle (x + ?) (x + ?)$, where the question marks are replaced with integers whose product is 12 and whose sum is $\displaystyle -7$; these integers are $\displaystyle -3\ and -4$.

$\displaystyle (u-3) (u-4)= 0$

Set each factor to zero and solve for $\displaystyle u$; then substitute back and solve for $\displaystyle x$:

$\displaystyle u-3 = 0$

$\displaystyle u=3$

$\displaystyle \sqrt{x} =3$

$\displaystyle x = 9$

$\displaystyle u-4 = 0$

$\displaystyle u=4$

$\displaystyle \sqrt{x} =4$

$\displaystyle x = 16$

The solution set, which can be confirmed by substitution, is $\displaystyle \left \{ 9,16\right \}$.

This can be best solved by substituting $\displaystyle u = \frac{1}{x}$ , and, subsequently, $\displaystyle u ^{2}=\left ( \frac{1}{x} \right ) ^{2} =\frac{1}{x ^{2}}$, then solving the resulting quadratic equation.

$\displaystyle \frac{1}{x^{2}} - \frac{8}{x} + 12 = 0$

$\displaystyle \frac{1}{x^{2}} - 8 \cdot \frac{1}{x} + 12 = 0$

$\displaystyle u^{2} - 8u + 12 = 0$

Factor the expression on the left by finding two integers whose product is 12 and whose sum is $\displaystyle -8$:

$\displaystyle (u-6)(u-2) = 0$

Set each linear binomial factor to 0, solve separately for $\displaystyle u$, and substitute back:

$\displaystyle u-6 = 0$

$\displaystyle u=6$

$\displaystyle \frac{1}{x} = 6$

$\displaystyle x = \frac{1}{6}$

or

$\displaystyle u-2 = 0$

$\displaystyle u=2$

$\displaystyle \frac{1}{x} = 2$

$\displaystyle x = \frac{1}{2}$

The variation equation for this situation is 

$\displaystyle \frac{P_{1}}{\sqrt{L_{1}}} = \frac{P_{2}}{\sqrt{L}_{2}}$

Set $\displaystyle P_{1} = 8.5 , L_{1} = 18, {L}_{2}= 21$, and solve for $\displaystyle P_{2}$;

$\displaystyle \frac{8.5}{\sqrt{18}} = \frac{P_{2}}{\sqrt{21}}$

$\displaystyle \frac{8.5}{\sqrt{18}} \cdot{\sqrt{21}} = \frac{P_{2}}{\sqrt{21}} \cdot{\sqrt{21}}$

$\displaystyle P_{2} = \frac{8.5}{\sqrt{18}} \cdot{\sqrt{21}} \approx 9.2 \textrm{ sec}$

$\displaystyle 2^{5} = 32$

$\displaystyle 2 = \sqrt[5]{32}$

$\displaystyle 8 = 2^{3} = \left ( \sqrt[5]{32} \right ) ^{3} = 32^{ \frac{3}{5} }$

$\displaystyle x = 8^{N}= \left (32^{ \frac{3}{5} } \right$^ {N } =32^{ \frac{3}{5} N }\)

$\displaystyle \log_{32} x =\log_{32}32^{ \frac{3}{5} N } = \frac{3}{5} N$

First, isolate the absolute value expression on one side:

$\displaystyle | 4x - 18 | - 18 = 57$

$\displaystyle | 4x - 18 | -18+ 18 = 57+18$

$\displaystyle | 4x - 18 | = 75$

Rewrite as a compound sentence:

$\displaystyle 4x - 18 = - 75 \textrm{ or } 4x - 18 = 75$

Solve each separately:

$\displaystyle 4x - 18 = - 75$

$\displaystyle 4x - 18+18 = -75+18$

$\displaystyle 4x= - 57$

$\displaystyle 4x \div 4= - 57 \div 4$

$\displaystyle x= - 14.25$

or

$\displaystyle 4x - 18 = 75$

$\displaystyle 4x - 18+18 = 75+18$

$\displaystyle 4x= 93$

$\displaystyle 4x \div 4=93 \div 4$

$\displaystyle x = 23.25$

The solution set is $\displaystyle \left \{ -14.25, 23.25 \right \}$

First, isolate the absolute value expression on one side:

$\displaystyle | 4x - 18 | + 18 = 57$

$\displaystyle | 4x - 18 | + 18 -18 = 57-18$

$\displaystyle | 4x - 18 | = 39$

Rewrite as a compound sentence:

$\displaystyle 4x - 18 = - 39 \textrm{ or } 4x - 18 = 39$

Solve each separately:

$\displaystyle 4x - 18 = - 39$

$\displaystyle 4x - 18+18 = - 39+18$

$\displaystyle 4x= - 21$

$\displaystyle 4x \div 4= - 21 \div 4$

$\displaystyle x= - 5.25$

or

$\displaystyle 4x - 18 = 39$

$\displaystyle 4x - 18+18 = 39+18$

$\displaystyle 4x= 57$

$\displaystyle 4x \div 4=57 \div 4$

$\displaystyle x = 14.25$

The solution set is $\displaystyle \left \{ -5.25, 14.25\right \}$.

Write this as a compound equation and solve each separately.

$\displaystyle \left | x ^{2} - 2x \right | = 3x -6$

$\displaystyle x ^{2} - 2x = 3x -6 \textrm{ or }x ^{2} - 2x = -\left ( 3x -6 \right )$

$\displaystyle x ^{2} - 2x = 3x -6$

$\displaystyle x ^{2} - 2x - 3x + 6 = 3x -6 - 3x + 6$

$\displaystyle x ^{2} - 5x + 6 =0$

$\displaystyle (x-3) (x-2) =0$

$\displaystyle x - 3 = 0 \Rightarrow x = 3$

$\displaystyle x - 2 = 0 \Rightarrow x =2$

$\displaystyle x ^{2} - 2x = -\left ( 3x -6 \right )$

$\displaystyle x ^{2} - 2x = -3x +6$

$\displaystyle x ^{2} - 2x+ 3x - 6= -3x +6 + 3x - 6$

$\displaystyle x ^{2} +x - 6= 0$

$\displaystyle (x + 3) (x - 2)= 0$

$\displaystyle x + 3 = 0 \Rightarrow x = -3$

$\displaystyle x - 2 = 0 \Rightarrow x =2$

This gives us three possible solutions - $\displaystyle \left \{ -3, 2, 3\right \}$. We check all three.

$\displaystyle x = -3$

$\displaystyle \left | x ^{2} - 2x \right | = 3x -6$

$\displaystyle \left | \left ( -3\right ) ^{2} - 2 (-3) \right | = 3 (-3) -6$

$\displaystyle \left | \left9 - (-6) \right | = -9-6$

$\displaystyle \left |15| = -15$

$\displaystyle 15 = -15$

This is a false statement so we can eliminate $\displaystyle -3$ as a false "solution".

$\displaystyle x = 2$

$\displaystyle \left | x ^{2} - 2x \right | = 3x -6$

$\displaystyle \left |2^{2} - 2 \cdot 2 \right | = 2 \cdot 3 -6$

$\displaystyle \left | \left 4-4 \right | = 6-6$

$\displaystyle |0| = 0$

$\displaystyle 0=0$

2 is a solution.

$\displaystyle x = 3$

$\displaystyle \left | x ^{2} - 2x \right | = 3x -6$

$\displaystyle \left | 3 ^{2} - 2 \cdot 3 \right | = 3 \cdot 3 -6$

$\displaystyle \left | \left 9 - 6\right | = 9-6$

$\displaystyle \left |3 | = 3$

$\displaystyle 3= 3$

3 is a solution.

The solution set is $\displaystyle x \in \left \{ 2, 3\right \}$.

Since $\displaystyle 2 ^{2} = 4$ and $\displaystyle 2^{3} = 8$, replace, and use the exponent rules:

$\displaystyle 4 ^{5x - 7} = 8 ^{2x -1}$

$\displaystyle \left ( 2 ^{2} \right ) ^{5x - 7} = \left ( 2^{3} \right ) ^{2x -1}$

$\displaystyle 2 ^{2 \left ( 5x - 7 \right )}=2 ^{3 \left ( 2x - 1 \right )}$

Set the exponents equal to each other and solve for $\displaystyle x$:

$\displaystyle 2 \left ( 5x - 7 \right )=3 \left ( 2x - 1 \right )$

$\displaystyle 10x -14=6x - 3$

$\displaystyle 10x-6x -14+14=6x -6x - 3+14$

$\displaystyle 4x = 11$

$\displaystyle 4x \div 4 = 11 \div 4$

$\displaystyle x = 2.75$