Question
Real Tutor Solution
Tutor-Verified Answer
Answer
\( v = 21 \)
Solution
To solve the equation:
\[
-1458 = -2(2v + 39)^{\frac{3}{2}}
\]
Follow these steps:
1. **Eliminate the Negative Signs:**
Multiply both sides by \(-1\) to simplify:
\[
1458 = 2(2v + 39)^{\frac{3}{2}}
\]
2. **Isolate the Exponential Term:**
Divide both sides by \(2\):
\[
729 = (2v + 39)^{\frac{3}{2}}
\]
3. **Remove the Exponent:**
To eliminate the exponent \(\frac{3}{2}\), raise both sides to the power of \(\frac{2}{3}\):
\[
729^{\frac{2}{3}} = 2v + 39
\]
**Simplify \(729^{\frac{2}{3}}\):**
- Recognize that \(729 = 9^3\).
- Therefore:
\[
729^{\frac{2}{3}} = (9^3)^{\frac{2}{3}} = 9^{2} = 81
\]
So,
\[
81 = 2v + 39
\]
4. **Solve for \(v\):**
Subtract \(39\) from both sides:
\[
81 - 39 = 2v \\
42 = 2v
\]
Divide by \(2\):
\[
v = 21
\]
**Final Answer:**
\(v = 21\)
Reviewed and approved by the UpStudy tutoring team
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Simplify this solution Extra Insights
To unravel this equation, let’s first isolate the expression involving \( v \). Start by dividing both sides by \(-2\): \[ 729 = (2v + 39)^{\frac{3}{2}} \] Next, to eliminate the power of \(\frac{3}{2}\), raise both sides to the \(\frac{2}{3}\) power: \[ (2v + 39) = 729^{\frac{2}{3}} \] Calculating \( 729^{\frac{2}{3}} \) gives us \( 27^2 = 729 \). Now, solving for \( v \): \[ 2v + 39 = 27 \] \[ 2v = 27 - 39 \] \[ 2v = -12 \] \[ v = -6 \] So, the solution to the equation is \( v = -6 \).
