Actually: \( h^{5/2} = 498.816 - 9.17 = 489.65 \) → \( h = (489.65)^{2/5} \).

["Understanding the Calculation: How to Solve ( h^{5/2} = 489.65 ) for ( h )", "When faced with equations involving fractional exponents like ( h^{5/2} = 489.65 ), solving for ( h ) requires understanding the algebraic transformation of exponents and careful computation. This article explains step-by-step how to compute ( h = (489.65)^{2/5} ), providing clarity for those tackling similar mathematical expressions.", "---", "### Breaking Down the Equation", "The original equation is:", "[\nh^{5/2} = 489.65\n]", "The exponent ( \frac{5}{2} ) means we are raising ( h ) to the power of ( 2.5 ), or equivalently, computing ( h ) raised to the fifth power, then taking the square root, or vice versa:", "[\nh^{5/2} = \left( h^{1/2} \right)^5 = \left( \sqrt{h} \right)^5\n]", "To isolate ( h ), we need to eliminate the fractional exponent ( \frac{5}{2} ). This is done by raising both sides of the equation to the reciprocal exponent, ( \frac{2}{5} ):", "[\nh = \left( h^{5/2} \right)^{2/5}\n]", "---", "### Step-by-Step Computation", "Given:\n[\nh = (489.65)^{2/5}\n]", "This expresses ( h ) as the number ( 489.65 ) raised to the ( \frac{2}{5} ) power.", "#### Step 1: Confirm the exponent interpretation", "Rewriting ( \left(489.65\right)^{2/5} ) explicitly:", "- Take the fifth root of ( 489.65 ):\n [\n \sqrt[5]{489.65} \approx x \quad \ ext{such that } x^5 = 489.65\n ]\n- Then square the result:\n [\n h = x^2\n ]", "#### Step 2: Deriving numerically", "Direct evaluation of ( (489.65)^{2/5} ) is computationally efficient using logarithms or a scientific calculator:", "Using logarithms:", "[\n\log h = \frac{2}{5} \log(489.65)\n]", "First, compute:", "[\n\log(489.65) \approx 2.6892\n]", "Then:", "[\n\log h = \frac{2}{5} \ imes 2.6892 = 1.07568\n]", "Now compute ( 10^{1.07568} ):", "[\n10^{1.07568} \approx 11.89 \ imes 10^{-1} = 11.89 \quad (\ ext{Wait: correction below})\n]", "Actually, use calculator precision:", "[\n\log(489.65) \approx 2.6892\n]", "[\n\frac{2}{5} \ imes 2.6892 = 1.07568\n]", "[\n10^{1.07568} \approx 11.89 \quad \ ext{is incorrect — recompute}\n]", "Actually, ( 10^{1.07568} = 10^{1 + 0.07568} = 10 \ imes 10^{0.07568} )", "[\n10^{0.07568} \approx 1.188 \quad \Rightarrow \quad 10 \ imes 1.188 = 11.88\n]", "Wait — this contradicts intuition. Let’s double-check with direct computation.", "Alternatively, verify via exponential:", "[\nh = (489.65)^{0.4} \quad (0.4 = \frac{2}{5})\n]", "Using a calculator:", "[\n489.65^{0.4} \approx 11.88\n]", "Thus,", "[\nh \approx 11.88\n]", "Wait — earlier numeric guess was off. Let's use accurate computation:", "### Accurate Numerical Calculation:", "[\nh = 489.65^{2/5}\n]", "Input in calculator:", "- ( 489.65^{0.4} \approx 11.88 )", "But this seems low. Let’s verify ( x = 12 ):", "[\n12^{5/2} = \left( \sqrt{12} \right)^5 \approx (3.464)^5 \approx 3.464^2 = 12 \Rightarrow 12^{2.5} = 12^2 \ imes \sqrt{12} = 144 \ imes 3.464 \approx 498.8\n]", "Getting close to 498.816.", "Try ( h^{5/2} = 489.65 ):", "Try ( h = 12^2 = 144 )? Not matching powers.", "Wait — step back.", "Actually, let's reverse:", "Given ( h^{5/2} = 489.65 )", "Try ( h = 144 ):\n( \sqrt{144} = 12 ), ( 12^5 = 248,832 ) → way too big.", "Mistake in size estimation.", "Let’s correctly evaluate ( x = 489.65^{2/5} )", "---", "### Correct Evaluation Using Logs and Precision", "[\n\log(489.65) = \log(489.65) \approx 2.6889 \quad \ ext{(using precise log table or calculator)}\n]", "Then:", "[\n\frac{2}{5} \ imes 2.6889 = 1.07556\n]", "Now compute:", "[\n10^{1.07556} = 10^{1} \ imes 10^{0.07556}\n]", "[\n10^{0.07556} \approx 1.1895 \quad (\ ext{since } 10^{0.07556} = e^{0.07556 \ln 10} \approx e^{0.07556 \ imes 2.3026} \approx e^{0.1738} \approx 1.1895)\n]", "Thus:", "[\nh \approx 10 \ imes 1.1895 = 11.895\n]", "But this still seems inconsistent with the given ( h^{5/2} = 489.65 )", "Wait — this indicates that ( h^{5/2} = 489.65 ) implies ( h \approx 11.9 ) is correct?", "Let’s compute ( (11.9)^{5/2} ):", "[\n\sqrt{11.9} \approx 3.646\n]", "[\n3.646^5 = (3.646^2)^2 \ imes 3.646 = (13.29)^2 \ imes 3.646 \approx 176.6 \ imes 3.646 \approx 643 \quad \ ext{Too low}\n]", "No — error in exponent interpretation.", "Wait: ( h^{5/2} = 489.65 \Rightarrow h = (489.65)^{2/5} )", "But ( 489.65 ) is very close to what we got earlier.", "Try:", "Let ( x = 489.65^{2/5} )", "Set ( y = x^{1/5} ), then ( y^5 = 489.65 )", "Try ( y = 3 ): ( 3^5 = 243 ) — too low\n( y = 4 ): ( 4^5 = 1024 ) — too high\n( y = 3.5 ): ( 3.5^2 = 12.25, 3.5^4 = 150.06, 3.5^5 = 525.2 ) — higher\n( y = 3.4 ): ( 3.4^2 = 11.56, 3.4^4 = 133.63, 3.4^5 = 455.26 ) — close\n( y = 3.42 ): ( 3.42^2 = 11.6964, ^4 = 136.68, ^5 = 11.6964 \ imes 136.68 \approx 1599? Wait — miscalc", "Better:", "( 3.42^5 = (3.42^2)^2 \cdot 3.42 = (11.6964)^2 \cdot 3.42 \approx 136.68 \ imes 3.42 \approx 467.3 )", "Still low.", "Try ( y = 3.45 ):\n( y^2 = 11.9025 ),\n( y^4 = (11.9025)^2 \approx 141.67 ),\n( y^5 = 141.67 \ imes 3.45 \approx 489.7 \approx 489.65 )", "Perfect.", "Thus:", "[\ny \approx 3.45 \Rightarrow h = y \approx 3.45\n]", "Wait — this contradicts prior computation.", "Resolution: we made sign error in exponentiation.", "Recall:", "From ( h = (489.65)^{2/5} )", "And ( 489.65^{2/5} = (489.65)^{0.4} )", "Compute ( 489.65^{0.4} ):", "Using ( \log_{10}(489.65) = \log_{10}(489.65) \approx 2.6888 )", "Then:", "[\n0.4 \ imes 2.6888 = 1.07552\n]", "[\n10^{1.07552} = 10^{1 + 0.07552} = 10 \ imes 10^{0.07552}\n]", "[\n10^{0.07552} = e^{0.07552 \ imes \ln 10} = e^{0.07552 \ imes 2.302585} = e^{0.17382} \approx 1.1893\n]", "Thus:", "[\nh \approx 10 \ imes 1.1893 = 11.893\n]", "But now ( h^{5/2} = (11.893)^{2.5} )", "Compute:", "( \sqrt{11.893} \approx 3.45 )", "( 3.45^2 = 11.8925 )\n( 3.45^4 = (11.8925)^2 \approx 141.66 )\n( 3.45^5 = 141.66 \ imes 3.45 \approx 489.5 )", "Yes — so ( (11.893)^{2.5} \approx 489.5 ), very close to 489.65.", "Thus, accurate solution is:", "[\nh \approx 11.89\n]", "But earlier step said ( h = (489.65)^{2/5} ), so:", "[\nh = 489.65^{2/5} = \left(489.65^{1/5}\right)^2\n]", "Compute ( 489.65^{1/5} ):", "[\n\log(489.65) = 2.6888, \quad \frac{1}{5} \ imes 2.6888 = 0.53776, "]