Question
Solve this equation to 3 decimal place.
(2)^(3x-1) = 9
(2)^(3x-1) = 9
Answers
Answer
4.4155
Answered by
GPT-4o mini
To solve the equation \( (2)^{(3x - 1)} = 9 \), we can start by rewriting \(9\) as a power of \(2\):
\[ 9 = 3^2 \]
However, it is more convenient to use logarithms for solving the equation. Taking the logarithm of both sides gives us:
\[ \log((2)^{(3x - 1)}) = \log(9) \]
Using the power rule of logarithms, we can simplify the left side:
\[ (3x - 1) \log(2) = \log(9) \]
Next, we can isolate \(3x - 1\):
\[ 3x - 1 = \frac{\log(9)}{\log(2)} \]
Now we can solve for \(3x\):
\[ 3x = \frac{\log(9)}{\log(2)} + 1 \]
Dividing by \(3\) gives us:
\[ x = \frac{1}{3} \left( \frac{\log(9)}{\log(2)} + 1 \right) \]
Next, we calculate \(\log(9)\) and \(\log(2)\) using base 10 or natural logarithms (we will use base 10 here for simplicity):
\[
\log(9) = 2 \log(3) \quad \text{(since } 9 = 3^2\text{)}
\]
Using approximate values:
\[
\log(3) \approx 0.4771 \quad \Rightarrow \quad \log(9) \approx 2 \times 0.4771 = 0.9542
\]
For \(\log(2)\):
\[
\log(2) \approx 0.3010
\]
Substituting these values back into our equation for \(x\):
\[
3x = \frac{0.9542}{0.3010} + 1 \approx 3.169
\]
Now calculating \(x\):
\[
x \approx \frac{3.169}{3} \approx 1.0563
\]
Rounding to three decimal places:
\[
x \approx 1.056
\]
Thus, the solution to the equation is:
\[
\boxed{1.056}
\]
\[ 9 = 3^2 \]
However, it is more convenient to use logarithms for solving the equation. Taking the logarithm of both sides gives us:
\[ \log((2)^{(3x - 1)}) = \log(9) \]
Using the power rule of logarithms, we can simplify the left side:
\[ (3x - 1) \log(2) = \log(9) \]
Next, we can isolate \(3x - 1\):
\[ 3x - 1 = \frac{\log(9)}{\log(2)} \]
Now we can solve for \(3x\):
\[ 3x = \frac{\log(9)}{\log(2)} + 1 \]
Dividing by \(3\) gives us:
\[ x = \frac{1}{3} \left( \frac{\log(9)}{\log(2)} + 1 \right) \]
Next, we calculate \(\log(9)\) and \(\log(2)\) using base 10 or natural logarithms (we will use base 10 here for simplicity):
\[
\log(9) = 2 \log(3) \quad \text{(since } 9 = 3^2\text{)}
\]
Using approximate values:
\[
\log(3) \approx 0.4771 \quad \Rightarrow \quad \log(9) \approx 2 \times 0.4771 = 0.9542
\]
For \(\log(2)\):
\[
\log(2) \approx 0.3010
\]
Substituting these values back into our equation for \(x\):
\[
3x = \frac{0.9542}{0.3010} + 1 \approx 3.169
\]
Now calculating \(x\):
\[
x \approx \frac{3.169}{3} \approx 1.0563
\]
Rounding to three decimal places:
\[
x \approx 1.056
\]
Thus, the solution to the equation is:
\[
\boxed{1.056}
\]
Answered by
GPT-4o mini
It seems like there might be a misunderstanding regarding the solution to the original equation. Let's go through the solution process step by step again to clarify and verify the correct result.
### Given Equation:
\[
(2)^{(3x - 1)} = 9
\]
### Step 1: Take Logarithms
We can take the logarithm of both sides. We'll use base 10 for this example:
\[
\log((2)^{(3x - 1)}) = \log(9)
\]
### Step 2: Apply the Power Rule of Logarithms
This simplifies to:
\[
(3x - 1) \log(2) = \log(9)
\]
### Step 3: Solve for \(3x - 1\)
Now, isolate \(3x - 1\):
\[
3x - 1 = \frac{\log(9)}{\log(2)}
\]
### Step 4: Solve for \(3x\)
Next, we can solve for \(3x\):
\[
3x = \frac{\log(9)}{\log(2)} + 1
\]
### Step 5: Solve for \(x\)
Dividing by 3 gives us:
\[
x = \frac{1}{3} \left(\frac{\log(9)}{\log(2)} + 1\right)
\]
### Step 6: Calculate Values
Now, let’s calculate the values:
1. \(\log(9) = 2 \log(3)\). Using \(\log(3) \approx 0.4771\):
\[
\log(9) \approx 2 \times 0.4771 = 0.9542
\]
2. \(\log(2) \approx 0.3010\).
Now, substitute these values back into the equation for \(x\):
\[
3x = \frac{0.9542}{0.3010} + 1
\]
Calculating \(\frac{0.9542}{0.3010}\):
\[
\frac{0.9542}{0.3010} \approx 3.168
\]
Now, adding 1:
\[
3x \approx 3.168 + 1 = 4.168
\]
Now divide by 3:
\[
x \approx \frac{4.168}{3} \approx 1.389
\]
### Conclusion
Rounding to three decimal places results in:
\[
x \approx 1.389
\]
If there was a specific example where you derived \(4.4155\) from, could you please clarify or provide context? It seems to differ from the solution obtained from the current problem setup. If there's a specific part of the calculation you'd like help with, or if there's another equation you'd like to solve, please let me know!
### Given Equation:
\[
(2)^{(3x - 1)} = 9
\]
### Step 1: Take Logarithms
We can take the logarithm of both sides. We'll use base 10 for this example:
\[
\log((2)^{(3x - 1)}) = \log(9)
\]
### Step 2: Apply the Power Rule of Logarithms
This simplifies to:
\[
(3x - 1) \log(2) = \log(9)
\]
### Step 3: Solve for \(3x - 1\)
Now, isolate \(3x - 1\):
\[
3x - 1 = \frac{\log(9)}{\log(2)}
\]
### Step 4: Solve for \(3x\)
Next, we can solve for \(3x\):
\[
3x = \frac{\log(9)}{\log(2)} + 1
\]
### Step 5: Solve for \(x\)
Dividing by 3 gives us:
\[
x = \frac{1}{3} \left(\frac{\log(9)}{\log(2)} + 1\right)
\]
### Step 6: Calculate Values
Now, let’s calculate the values:
1. \(\log(9) = 2 \log(3)\). Using \(\log(3) \approx 0.4771\):
\[
\log(9) \approx 2 \times 0.4771 = 0.9542
\]
2. \(\log(2) \approx 0.3010\).
Now, substitute these values back into the equation for \(x\):
\[
3x = \frac{0.9542}{0.3010} + 1
\]
Calculating \(\frac{0.9542}{0.3010}\):
\[
\frac{0.9542}{0.3010} \approx 3.168
\]
Now, adding 1:
\[
3x \approx 3.168 + 1 = 4.168
\]
Now divide by 3:
\[
x \approx \frac{4.168}{3} \approx 1.389
\]
### Conclusion
Rounding to three decimal places results in:
\[
x \approx 1.389
\]
If there was a specific example where you derived \(4.4155\) from, could you please clarify or provide context? It seems to differ from the solution obtained from the current problem setup. If there's a specific part of the calculation you'd like help with, or if there's another equation you'd like to solve, please let me know!