Questions: The equation (2^x+1=19) can be solved exactly and written in the form of (fraclog a blog c-d), where (a, b, c), and (d) represent digits. Place the value of (a) in the first box. Place the value of (b) in the second box. Place the value of (c) in the third box. Place the value of (d) in the fourth box. 1 9 2 1

Solution Steps

To solve the equation \(2^{x+1} = 19\) and express \(x\) in the form \(\frac{\log a b}{\log c} - d\), we can follow these steps:

1. Take the logarithm of both sides of the equation to solve for \(x\).
2. Use properties of logarithms to isolate \(x\).
3. Identify the values of \(a\), \(b\), \(c\), and \(d\) from the resulting expression.

1. Take the logarithm of both sides: \(\log(2^{x+1}) = \log(19)\).
2. Use the power rule of logarithms: \((x+1) \log(2) = \log(19)\).
3. Solve for \(x\): \(x = \frac{\log(19)}{\log(2)} - 1\).
4. Identify \(a\), \(b\), \(c\), and \(d\) from the expression \(\frac{\log(19)}{\log(2)} - 1\).

Starting with the equation \(2^{x+1} = 19\), we take the logarithm of both sides:

\[ \log(2^{x+1}) = \log(19) \]

Using the power rule of logarithms, we can rewrite the left side:

\[ (x+1) \log(2) = \log(19) \]

Next, we isolate \(x\):

\[ x + 1 = \frac{\log(19)}{\log(2)} \]

Subtracting 1 from both sides gives:

\[ x = \frac{\log(19)}{\log(2)} - 1 \]

From the expression \(x = \frac{\log(1 \cdot 9)}{\log(2)} - 1\), we identify:

• \(a = 1\)
• \(b = 9\)
• \(c = 2\)
• \(d = 1\)

Substituting the values into the expression, we find:

\[ x \approx 2.1699 \]

Final Answer