Questions: Solve the equation by rewriting the exponential expression using the indicated logarithm. Then use a calculator to approximate the variable to 3 decimal places. 4^(3x-1)=39 using the common log. x=

Solution Steps

To solve the equation \(4^{3x-1} = 39\) using the common logarithm (base 10), we can follow these steps:

1. Take the common logarithm of both sides of the equation.
2. Use the logarithm power rule to bring the exponent in front of the logarithm.
3. Solve for \(x\) by isolating it on one side of the equation.
4. Use a calculator to approximate the value of \(x\) to three decimal places.

We start with the equation:

\[ 4^{3x - 1} = 39 \]

Taking the common logarithm of both sides gives us:

\[ \log(4^{3x - 1}) = \log(39) \]

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

\[ (3x - 1) \cdot \log(4) = \log(39) \]

Next, we isolate \(3x - 1\):

\[ 3x - 1 = \frac{\log(39)}{\log(4)} \]

Now, we can solve for \(x\):

\[ 3x = \frac{\log(39)}{\log(4)} + 1 \]

\[ x = \frac{\frac{\log(39)}{\log(4)} + 1}{3} \]

Using the values:

\[ \log(4) \approx 0.6021 \quad \text{and} \quad \log(39) \approx 1.5911 \]

We substitute these into the equation:

\[ x = \frac{\frac{1.5911}{0.6021} + 1}{3} \]

Calculating this gives:

\[ x \approx \frac{2.6442}{3} \approx 1.2142 \]

Final Answer