7(3^x)=7^y,and x^2-y+1=0?
回答 (2)
2015-12-25 2:41 pm
✔ 最佳答案
7(3ˣ) = 7ʸ ⇒ 3ˣ = 7ʸ⁻¹ and x² - y + 1 = 0 ⇒ y - 1 = x² , so 3ˣ = 7^x²
3ˣ = (7ˣ)ˣ
3 = 7ˣ
x = log3 / log7 = 0.564575034
y = x²+1 = (log3 / log7)² + 1 = 1.318744969
2015-12-26 9:37 am
7(3ˣ) = 7ʸ ⇒ 3ˣ = 7ʸ⁻¹ and x² - y + 1 = 0 ⇒ y - 1 = x² , so
3ˣ = 7^x²
3ˣ = (7ˣ)ˣ
3 = 7ˣ
x = log3 / log7 = 0.564575034
y = x²+1 = (log3 / log7)² + 1 = 1.318744969
3ˣ = 7^x²
3ˣ = (7ˣ)ˣ
3 = 7ˣ
x = log3 / log7 = 0.564575034
y = x²+1 = (log3 / log7)² + 1 = 1.318744969
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