Module: Foundation Mathematics I (ITE3705)

Unit: GCNU303A – Solve problems related to simple exponential and logarithmic functions

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Questions / 題目

Section 1: Laws of Rational Indices / 有理指數定律

Question 1

A player is farming craft materials in an anime-style mobile RPG. The total inventory capacity is modeled by ${\displaystyle \frac{(x^4{y}^{-2}{)}^3}{x^5{y}^{-8}}}$x5y−8(x4y−2)3​. Simplify the expression and express your answer with positive indices.

一名玩家正在一款動漫風格的手遊中收集鍛造材料。其道具背包容量以 ${\displaystyle \frac{(x^4{y}^{-2}{)}^3}{x^5{y}^{-8}}}$x5y−8(x4y−2)3​ 表示。請化簡此算式，並以正指數表示答案。

Question 2

Solve the index equation: $9^{x-1}={\left({\displaystyle \frac{1}{27}}\right)}^{x+1}$9x−1=(271​)x+1.

解指數方程：$9^{x-1}={\left({\displaystyle \frac{1}{27}}\right)}^{x+1}$9x−1=(271​)x+1。

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Section 2: Exponential Equations / 指數方程

Question 3

Solve the exponential equation: $3^x\times 9^{x+2}=243$3x×9x+2=243.

解指數方程：$3^x\times 9^{x+2}=243$3x×9x+2=243。

Question 4

In a fantasy battle game, the drop rate multiplier $x$x satisfies the equation: $4^{2x-1}={\left({\displaystyle \frac{1}{32}}\right)}^{x-3}$42x−1=(321​)x−3. Find the value of $x$x.

在一款奇幻戰鬥遊戲中，掉寶率加成系數 $x$x 符合方程：$4^{2x-1}={\left({\displaystyle \frac{1}{32}}\right)}^{x-3}$42x−1=(321​)x−3。求 $x$x 的值。

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Section 3: Applications of Exponential Functions / 指數函數的應用

Question 5

An IVE graduate deposits HK$50,000 into a Hong Kong bank fixed-deposit account that offers an interest rate of 4% per annum. Find the total amount after 3 years if the interest is compounded monthly. (Give the answer correct to the nearest integer.)

一名 IVE 畢業生將 50,000 港元存入香港某銀行的定期存款戶口，年利率為 4%。若每月結息一次（每月複利），求 3 年後的本利和。（答案須準確至最接近的整數。）

Question 6

During a limited-time raid event on a Hong Kong gaming server, monster activity $Q$Q increases exponentially according to $Q(t)=Q_0{e}^{kt}$Q(t)=Q0​ekt. Initially, there are 20,000 monsters, and the count grows continuously at a rate of 5% per hour. Find the number of monsters after 6 hours. (Give the answer correct to the nearest integer.)

在香港伺服器的限時討伐活動中，魔物活躍數量 $Q$Q 依指數規律 $Q(t)=Q_0{e}^{kt}$Q(t)=Q0​ekt 增長。初始魔物數量為 20,000 隻，並以每小時 5% 的速率持續增長。求 6 小時後的魔物數量。（答案須準確至最接近的整數。）

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Section 4: Logarithmic Equations & Properties / 對數方程與性質

Question 7

Simplify the expression without using a calculator: ${\log }_{10}25+2{\log }_{10}2-{\log }_{10}1$log10​25+2log10​2−log10​1.

在不使用計算機的情況下，化簡算式：${\log }_{10}25+2{\log }_{10}2-{\log }_{10}1$log10​25+2log10​2−log10​1。

Question 8

Solve the logarithmic equation: ${\log }_2(x+3)+{\log }_2(x-3)=4$log2​(x+3)+log2​(x−3)=4.

解對數方程：${\log }_2(x+3)+{\log }_2(x-3)=4$log2​(x+3)+log2​(x−3)=4。

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Section 5: Applications of Logarithmic Functions / 對數函數的應用

Question 9

A commuter loads HK$1,000 into a smart Octopus card investment plan that yields 5% continuous annual growth ($A=P{e}^{rt}$A=Pert). How many years will it take for the stored value to double? (Give your answer correct to 1 decimal place.)

一名通勤者將 1,000 港元充值至具備 5% 年連續增長率的八達通理財計劃中（$A=P{e}^{rt}$A=Pert）。儲值額需要多少年才能翻倍？（答案須準確至小數點後 1 位。）

Question 10

A trendy smartphone bought in Sham Shui Po costs HK$8,800 when brand-new. It depreciates steadily by 18% per year ($V=8800(1-0.18{)}^n$V=8800(1−0.18)n)[cite: 1]. After how many years will its trade-in value drop below HK$3,000?

一部在深水埗購買的潮流智能手機全新售價為 8,800 港元。該手機每年穩定貶值 18%（$V=8800(1-0.18{)}^n$V=8800(1−0.18)n）。多少年後，它的回收二手價會低於 3,000 港元？

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Solutions / 參考答案與題解

Solution 1

Simplify powers inside the numerator first: $(x^4{y}^{-2}{)}^3=x^{4\times 3}{y}^{-2\times 3}=x^{12}{y}^{-6}$(x4y−2)3=x4×3y−2×3=x12y−6.

首先展開分子括號：$(x^4{y}^{-2}{)}^3=x^{4\times 3}{y}^{-2\times 3}=x^{12}{y}^{-6}$(x4y−2)3=x4×3y−2×3=x12y−6。

Divide the numerator by the denominator: ${\displaystyle \frac{x^{12}{y}^{-6}}{x^5{y}^{-8}}}=x^{12-5}{y}^{-6-(-8)}=x^7{y}^2$x5y−8x12y−6​=x12−5y−6−(−8)=x7y2.

將分子除以分母：${\displaystyle \frac{x^{12}{y}^{-6}}{x^5{y}^{-8}}}=x^{12-5}{y}^{-6-(-8)}=x^7{y}^2$x5y−8x12y−6​=x12−5y−6−(−8)=x7y2。$${\mathbf{x}}^{\mathbf{7}}{\mathbf{y}}^{\mathbf{2}}$$x7y2

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Solution 2

Express both sides using base 3: $(3^2{)}^{x-1}=(3^{-3}{)}^{x+1}$(32)x−1=(3−3)x+1.

將兩邊轉化為以 3 為底數的表示式：$(3^2{)}^{x-1}=(3^{-3}{)}^{x+1}$(32)x−1=(3−3)x+1。

Multiply indices: $3^{2x-2}=3^{-3x-3}$32x−2=3−3x−3.

指數相乘：$3^{2x-2}=3^{-3x-3}$32x−2=3−3x−3。

Equate the indices: $2x-2=-3x-3$2x−2=−3x−3.

底數相同，指數相等：$2x-2=-3x-3$2x−2=−3x−3。$$5x=-1⟹\mathbf{x}=-{\displaystyle \frac{\mathbf{1}}{\mathbf{5}}}(\text{or }-0.2)$$5x=−1⟹x=−51​(or −0.2)

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Solution 3

Express all terms with base 3: $3^x\times (3^2{)}^{x+2}=3^5$3x×(32)x+2=35.

將所有項轉為底數 3：$3^x\times (3^2{)}^{x+2}=3^5$3x×(32)x+2=35。

Combine indices on the left-hand side: $3^{x+2x+4}=3^5⟹3^{3x+4}=3^5$3x+2x+4=35⟹33x+4=35.

合併左邊的指數：$3^{x+2x+4}=3^5⟹3^{3x+4}=3^5$3x+2x+4=35⟹33x+4=35。

Equate indices: $3x+4=5⟹3x=1⟹\mathbf{x}={\displaystyle \frac{\mathbf{1}}{\mathbf{3}}}$3x+4=5⟹3x=1⟹x=31​.

比較指數求解：$3x+4=5⟹3x=1⟹\mathbf{x}={\displaystyle \frac{\mathbf{1}}{\mathbf{3}}}$3x+4=5⟹3x=1⟹x=31​。

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Solution 4

Express base 4 and base 32 in base 2: $(2^2{)}^{2x-1}=(2^{-5}{)}^{x-3}$(22)2x−1=(2−5)x−3.

將底數 4 與 32 化為底數 2：$(2^2{)}^{2x-1}=(2^{-5}{)}^{x-3}$(22)2x−1=(2−5)x−3。

Simplify indices: $2^{4x-2}=2^{-5x+15}$24x−2=2−5x+15.

化簡指數：$2^{4x-2}=2^{-5x+15}$24x−2=2−5x+15。

Equate exponents: $4x-2=-5x+15⟹9x=17⟹\mathbf{x}={\displaystyle \frac{\mathbf{17}}{\mathbf{9}}}$4x−2=−5x+15⟹9x=17⟹x=917​.

建立指數等式：$4x-2=-5x+15⟹9x=17⟹\mathbf{x}={\displaystyle \frac{\mathbf{17}}{\mathbf{9}}}$4x−2=−5x+15⟹9x=17⟹x=917​。

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Solution 5

Identify compound interest variables: $P=50000$P=50000, $r\mathrm{\%}={\displaystyle \frac{4\mathrm{\%}}{12}}={\displaystyle \frac{0.04}{12}}$r%=124%​=120.04​, $n=3\times 12=36$n=3×12=36.

確認複利息各項數值：$P=50000$P=50000，$r\mathrm{\%}={\displaystyle \frac{4\mathrm{\%}}{12}}={\displaystyle \frac{0.04}{12}}$r%=124%​=120.04​，$n=3\times 12=36$n=3×12=36。

Substitute into $A=P{(1+{\displaystyle \frac{r}{m}})}^{mt}$A=P(1+mr​)mt: $A=50000{(1+{\displaystyle \frac{0.04}{12}})}^{36}$A=50000(1+120.04​)36.

代入公式 $A=P{(1+{\displaystyle \frac{r}{m}})}^{mt}$A=P(1+mr​)mt：$A=50000{(1+{\displaystyle \frac{0.04}{12}})}^{36}$A=50000(1+120.04​)36。

Calculate the final amount: $A\approx 50000\times 1.12727=\mathbf{H}\mathbf{K}\mathbf{\$}\mathbf{56},\mathbf{364}$A≈50000×1.12727=HK$56,364.

計算最終金額：$A\approx 50000\times 1.12727=\mathbf{56},\mathbf{364}\text{ 港元}$A≈50000×1.12727=56,364 港元。

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Solution 6

Use the continuous exponential growth formula: $Q(t)=Q_0{e}^{kt}$Q(t)=Q0​ekt, where $Q_0=20000$Q0​=20000, $k=0.05$k=0.05, $t=6$t=6.

應用連續指數增長公式：$Q(t)=Q_0{e}^{kt}$Q(t)=Q0​ekt，其中 $Q_0=20000$Q0​=20000，$k=0.05$k=0.05，$t=6$t=6。

Set up the equation: $Q(6)=20000\times e^{0.05\times 6}=20000\times e^{0.30}$Q(6)=20000×e0.05×6=20000×e0.30.

建立方程式：$Q(6)=20000\times e^{0.05\times 6}=20000\times e^{0.30}$Q(6)=20000×e0.05×6=20000×e0.30。

Evaluate: $Q(6)\approx 20000\times 1.34986=\mathbf{26},\mathbf{997}\text{ monsters / 隻魔物}$Q(6)≈20000×1.34986=26,997 monsters / 隻魔物.

數值計算：$Q(6)\approx 20000\times 1.34986=\mathbf{26},\mathbf{997}\text{ monsters / 隻魔物}$Q(6)≈20000×1.34986=26,997 monsters / 隻魔物。

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Solution 7

Apply power and zero logarithm laws: $2{\log }_{10}2={\log }_{10}(2^2)={\log }_{10}4$2log10​2=log10​(22)=log10​4 and ${\log }_{10}1=0$log10​1=0.

應用對數的冪次法則及零法則：$2{\log }_{10}2={\log }_{10}(2^2)={\log }_{10}4$2log10​2=log10​(22)=log10​4 且 ${\log }_{10}1=0$log10​1=0。

Combine terms: ${\log }_{10}25+{\log }_{10}4-0={\log }_{10}(25\times 4)={\log }_{10}(100)$log10​25+log10​4−0=log10​(25×4)=log10​(100).

結合對數項：${\log }_{10}25+{\log }_{10}4-0={\log }_{10}(25\times 4)={\log }_{10}(100)$log10​25+log10​4−0=log10​(25×4)=log10​(100)。

Since $100={10}^2$100=102: ${\log }_{10}({10}^2)=\mathbf{2}$log10​(102)=2.

因為 $100={10}^2$100=102：${\log }_{10}({10}^2)=\mathbf{2}$log10​(102)=2。

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Solution 8

Combine the logarithms: ${\log }_2[(x+3)(x-3)]=4⟹{\log }_2(x^2-9)=4$log2​[(x+3)(x−3)]=4⟹log2​(x2−9)=4.

合併左邊對數：${\log }_2[(x+3)(x-3)]=4⟹{\log }_2(x^2-9)=4$log2​[(x+3)(x−3)]=4⟹log2​(x2−9)=4。

Convert to exponential form: $x^2-9=2^4=16$x2−9=24=16.

轉化為指數形式：$x^2-9=2^4=16$x2−9=24=16。

Solve for $x$x: $x^2=25⟹x=5$x2=25⟹x=5 or $x=-5$x=−5.

解 $x$x：$x^2=25⟹x=5$x2=25⟹x=5 或 $x=-5$x=−5。

Check domain constraints: $x+3>0$x+3>0 and $x-3>0$x−3>0 require $x>3$x>3; hence reject $x=-5$x=−5.

檢驗真數定義域：由 $x+3>0$x+3>0 及 $x-3>0$x−3>0 得 $x>3$x>3，故捨去 $x=-5$x=−5。$$\mathbf{x}=\mathbf{5}$$x=5

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Solution 9

Set up the doubling equation: $2000=1000e^{0.05t}⟹e^{0.05t}=2$2000=1000e0.05t⟹e0.05t=2.

建立翻倍方程式：$2000=1000e^{0.05t}⟹e^{0.05t}=2$2000=1000e0.05t⟹e0.05t=2。

Take natural logarithms on both sides: $\ln (e^{0.05t})=\ln (2)⟹0.05t=\ln (2)$ln(e0.05t)=ln(2)⟹0.05t=ln(2).

兩邊取自然對數：$\ln (e^{0.05t})=\ln (2)⟹0.05t=\ln (2)$ln(e0.05t)=ln(2)⟹0.05t=ln(2)。

Calculate $t$t: $t={\displaystyle \frac{\ln 2}{0.05}}\approx {\displaystyle \frac{0.69315}{0.05}}\approx \mathbf{13.9}\text{ years / 年}$t=0.05ln2​≈0.050.69315​≈13.9 years / 年.

計算 $t$t：$t={\displaystyle \frac{\ln 2}{0.05}}\approx {\displaystyle \frac{0.69315}{0.05}}\approx \mathbf{13.9}\text{ years / 年}$t=0.05ln2​≈0.050.69315​≈13.9 years / 年。

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Solution 10

Set up the inequality: $8800(1-0.18{)}^n<3000⟹(0.82{)}^n<{\displaystyle \frac{3000}{8800}}\approx 0.3409$8800(1−0.18)n<3000⟹(0.82)n<88003000​≈0.3409.

列出不等式：$8800(1-0.18{)}^n<3000⟹(0.82{)}^n<{\displaystyle \frac{3000}{8800}}\approx 0.3409$8800(1−0.18)n<3000⟹(0.82)n<88003000​≈0.3409。

Take common logarithms on both sides: $\log ({0.82}^n)<\log (0.3409)⟹n\log (0.82)<\log (0.3409)$log(0.82n)<log(0.3409)⟹nlog(0.82)<log(0.3409).

兩邊取常用對數：$\log ({0.82}^n)<\log (0.3409)⟹n\log (0.82)<\log (0.3409)$log(0.82n)<log(0.3409)⟹nlog(0.82)<log(0.3409)。

Since $\log (0.82)\approx -0.08619<0$log(0.82)≈−0.08619<0, reverse the inequality sign: $n>{\displaystyle \frac{\log 0.3409}{\log 0.82}}\approx {\displaystyle \frac{-0.46737}{-0.08619}}\approx 5.42$n>log0.82log0.3409​≈−0.08619−0.46737​≈5.42.

由於 $\log (0.82)\approx -0.08619<0$log(0.82)≈−0.08619<0，除以負數時須轉變不等號方向：$n>{\displaystyle \frac{\log 0.3409}{\log 0.82}}\approx {\displaystyle \frac{-0.46737}{-0.08619}}\approx 5.42$n>log0.82log0.3409​≈−0.08619−0.46737​≈5.42。

Take the next integer: It will take 6 years for the value to drop below HK$3,000.

取下一位整數：即 6 年 後其回收價值會低於 3,000 港元。

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