题目
Problem
(a) Show that the equation log 3 ( 2 x + 1 ) = 1 + 2 log 3 ( x − 1 ) \log_3(2x+1)=1+2\log_3(x-1) log 3 ( 2 x + 1 ) = 1 + 2 log 3 ( x − 1 ) x x x
(3)
(b) Hence solve the equation log 3 ( 4 y + 1 ) = 1 + 2 log 3 ( 2 y − 1 ) \log_3(4y+1)=1+2\log_3(2y-1) log 3 ( 4 y + 1 ) = 1 + 2 log 3 ( 2 y − 1 )
(2)
题目中文翻译
(a) 证明方程 log 3 ( 2 x + 1 ) = 1 + 2 log 3 ( x − 1 ) \log_3(2x+1)=1+2\log_3(x-1) log 3 ( 2 x + 1 ) = 1 + 2 log 3 ( x − 1 ) x x x
(3)
(b) 由此解方程 log 3 ( 4 y + 1 ) = 1 + 2 log 3 ( 2 y − 1 ) \log_3(4y+1)=1+2\log_3(2y-1) log 3 ( 4 y + 1 ) = 1 + 2 log 3 ( 2 y − 1 )
(2)
原卷通用作答要求
Answer all questions. You should use a calculator where appropriate.
You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.
Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.
请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程；仅给出计算器答案而没有过程，不予评分。除非题目另有精度要求，非精确数值答案保留 3 位有效数字，以度为单位的角度保留 1 位小数。
解答
(a)
解法一
思路
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将常数 1 1 1 log 3 3 \log_3 3 log 3 3
答题过程
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The domain requires x > 1 x>1 x > 1
1 + 2 log 3 ( x − 1 ) = log 3 3 + log 3 ( x − 1 ) 2 = log 3 [ 3 ( x − 1 ) 2 ] . \begin{align*}
1+2\log_3(x-1)
=&\,\log_3 3+\log_3(x-1)^2\\[4mm]
=&\,\log_3\big[3(x-1)^2\big].
\end{align*} 1 + 2 log 3 ( x − 1 ) = = log 3 3 + log 3 ( x − 1 ) 2 log 3 [ 3 ( x − 1 ) 2 ] . Therefore
2 x + 1 = 3 ( x − 1 ) 2 . 2x+1=3(x-1)^2. 2 x + 1 = 3 ( x − 1 ) 2 . Expanding and rearranging,
2 x + 1 = 3 x 2 − 6 x + 3 , 2x+1=3x^2-6x+3, 2 x + 1 = 3 x 2 − 6 x + 3 , so the quadratic equation is
3 x 2 − 8 x + 2 = 0 . \boxed{3x^2-8x+2=0}. 3 x 2 − 8 x + 2 = 0 .
(b)
解法一
思路
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令 x = 2 y x=2y x = 2 y 2 2 2 y > 1 / 2 y>1/2 y > 1/2
答题过程
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Put x = 2 y x=2y x = 2 y
3 x 2 − 8 x + 2 = 0 , x = 8 ± 64 − 24 6 = 4 ± 10 3 . \begin{align*}
3x^2-8x+2=&\,0,\\[4mm]
x=&\,\frac{8\pm\sqrt{64-24}}6\\[4mm]
=&\,\frac{4\pm\sqrt{10}}3.
\end{align*} 3 x 2 − 8 x + 2 = x = = 0 , 6 8 ± 64 − 24 3 4 ± 10 . Thus
y = 4 ± 10 6 . y=\frac{4\pm\sqrt{10}}6. y = 6 4 ± 10 . The logarithm log 3 ( 2 y − 1 ) \log_3(2y-1) log 3 ( 2 y − 1 ) y > 1 / 2 y>1/2 y > 1/2
y = 4 + 10 6 ≈ 1.19 y=\frac{4+\sqrt{10}}6\approx\boxed{1.19} y = 6 4 + 10 ≈ 1.19 to 2 decimal places.
