Let a=32a = 3\sqrt{2}a=32 and b=151/66b = \dfrac{1}{5^{1/6}\sqrt{6}}b=51/661. If x,y∈Rx, y \in \mathbb{R}x,y∈R are such that
3x+2y=loga(18)54and3x + 2y = \log_a (18)^{\frac{5}{4}} \quad\text{and}3x+2y=loga(18)45and2x−y=logb(1080),2x - y = \log_b\left(\sqrt{1080}\right),2x−y=logb(1080),
then 4x+5y4x + 5y4x+5y is equal to _____.
Official answer: 8
First equation. a=32=181/2a = 3\sqrt2 = 18^{1/2}a=32=181/2, so loga185/4=5/41/2=52\log_a 18^{5/4} = \dfrac{5/4}{1/2} = \dfrac52loga185/4=1/25/4=25. Hence 3x+2y=523x + 2y = \dfrac523x+2y=25.
Second equation. 1080=23⋅33⋅51080 = 2^3\cdot3^3\cdot51080=23⋅33⋅5, so 1080=63/2 51/2\sqrt{1080} = 6^{3/2}\,5^{1/2}1080=63/251/2. Also b=5−1/6 6−1/2b = 5^{-1/6}\,6^{-1/2}b=5−1/66−1/2, so b−3=51/2 63/2=1080b^{-3} = 5^{1/2}\,6^{3/2} = \sqrt{1080}b−3=51/263/2=1080. Hence 2x−y=−32x - y = -32x−y=−3.
Solve. y=2x+3y = 2x + 3y=2x+3, so 3x+4x+6=523x + 4x + 6 = \frac523x+4x+6=25, giving x=−12x = -\frac12x=−21 and y=2y = 2y=2.
4x+5y=−2+10=84x + 5y = -2 + 10 = 84x+5y=−2+10=8.