\(\displaystyle 3\) is the only choice from those given that satisfies the equation. Substition of \(\displaystyle 3\) for \(\displaystyle x\) gives:

\(\displaystyle 4^3 = \frac{192}{3} \rightarrow 64 = 64\)

To solve for \(\displaystyle x\) in the equation \(\displaystyle 2x^2 - 32 =0\)

Factor \(\displaystyle 2\) out of the expression on the left of the equation:

\(\displaystyle \rightarrow 2(x^2 - 16) =0\)

Use the "difference of squares" technique to factor the parenthetical term on the left side of the equation.

\(\displaystyle \rightarrow 2(x+4)(x-4)=0\)

Any variable that causes any one of the parenthetical terms to become \(\displaystyle 0\) will be a valid solution for the equation. \(\displaystyle x+4\) becomes \(\displaystyle 0\) when \(\displaystyle x\) is \(\displaystyle -4\), and \(\displaystyle x-4\) becomes \(\displaystyle 0\) when \(\displaystyle x\) is \(\displaystyle 4\), so the solutions are \(\displaystyle -4\) and \(\displaystyle 4\).

Take the common logarithm of both sides and solve for \(\displaystyle y\):

\(\displaystyle 8^{y} = 100\)

\(\displaystyle \log 8^{y} = \log 100\)

\(\displaystyle y \log 8 =2\)

\(\displaystyle y =\frac{2}{\log 8} \approx \frac{2}{0.9031} \approx 2.21\)

\(\displaystyle \frac{1}{256} = \frac{1}{4 ^{4}} = 4 ^{-4}\), so \(\displaystyle 4 ^{2x - 7} = \frac{1}{256}\) can be rewritten as

\(\displaystyle 4 ^{2x - 7} = 4 ^{-4}\)

\(\displaystyle \log_{4} 4 ^{2x - 7} =\log_{4} 4 ^{-4}\)

\(\displaystyle 2x - 7 = -4\)

\(\displaystyle 2x - 7 + 7 = -4+ 7\)

\(\displaystyle 2x = 3\)

\(\displaystyle x = 1 .5\)

One method: Take the natural logarithm of both sides and solve for \(\displaystyle t\):

\(\displaystyle 80 ^{t} = 39\)

\(\displaystyle \ln 80 ^{t} = \ln 39\)

\(\displaystyle t \ln 80 = \ln 39\)

\(\displaystyle t =\frac{ \ln 39}{ \ln 80} \approx \frac{ 3.66}{ 4.38} \approx 0.84\)

Since \(\displaystyle 9 = 3^{2}\), we can rewrite this equation by subsituting and applying the power rule:

\(\displaystyle 3 ^{4x + 7} = 9 ^{2x +3}\)

\(\displaystyle 3 ^{4x + 7} = \left (3 ^{2} \right ) ^{2x +3}\)

\(\displaystyle 3 ^{4x + 7} =3 ^{2 \left ( 2x +3 \right )}\)

\(\displaystyle 3 ^{4x + 7} =3 ^{4x +6}\)

\(\displaystyle \log_{3} 3 ^{4x + 7} = \log_{3} 3 ^{4x +6}\)

\(\displaystyle 4x + 7 = 4x + 6\)

\(\displaystyle 4x + 7 -4x = 4x + 6-4x\)

\(\displaystyle 7 = 6\)

This statement is identically false, which means that the original equation is identically false. There is no solution.

\(\displaystyle \frac{1}{4} = \frac{1}{2^{2}} =2^{-2}\), so we can rewrite the equation as follows:

\(\displaystyle \left (\frac{1}{4} \right) ^{x - 7} = 2^{2x +5}\)

\(\displaystyle \left (2^{-2}) ^{x - 7} = 2^{2x +5}\)

\(\displaystyle 2^{-2 \left ( x - 7 \right )} = 2^{2x +5}\)

\(\displaystyle 2^{-2x +14 \right )} = 2^{2x +5}\)

\(\displaystyle \log_2 2^{-2x +14 \right )} =\log_2 2^{2x +5}\)

\(\displaystyle -2x + 14 = 2x + 5\)

\(\displaystyle -2x + 14 +2x -5= 2x + 5+2x -5\)

\(\displaystyle 9= 4x\)

\(\displaystyle x = 2.25\)

To find the y-intercepts, set \(\displaystyle x=0\) and solve.

\(\displaystyle y=\frac{3x^2-5}{2x^2+7}\)

\(\displaystyle y=\frac{3(0)^2-5}{2(0)^2+7}\)

\(\displaystyle y=\frac{-5}{7}\)