Review Exercise 2 — Class 9 Mathematics Solutions

Solution

1. Understand the exponent of 10.

   The given expression is $5.2\times {10}^6$. Since the power of $10$ is positive $6$, we multiply $5.2$ by ${10}^6=\mathrm{1,000,000}$, which shifts the decimal point $6$ places to the right.

2. Shift the decimal point.

   $$\begin{array}{cc}5.2\times {10}^6& =5.2\times \mathrm{1,000,000}\\ & =\mathrm{5,200,000}\end{array}$$

3. Identify the correct option.

   The value $\mathrm{5,200,000}$ corresponds to option (c).

Solution

1. Identify the standard form of scientific notation.

   Scientific notation is written in the form $A\times {10}^n$, where $1\le A<10$ and $n$ is an integer.

2. Move the decimal point to obtain $A$.

   For $0.00034$, we move the decimal point $4$ places to the right to place it after the first non-zero digit $3$, giving $A=3.4$.

3. Determine the power of 10.

   Since the decimal was moved $4$ places to the right (for a number strictly less than $1$), the power of $10$ is $-4$:

   $$\begin{array}{cc}0.00034& =3.4\times {10}^{-4}\end{array}$$

4. Identify the correct option.

   This matches option (b).

Solution

1. Recall the definition of common logarithm.

   Logarithms with base $10$ are known as common logarithms (or Briggian logarithms), denoted simply as $\log x$ or ${\log }_{10}x$.

2. Identify the correct option.

   Therefore, the base of a common logarithm is $10$, which corresponds to option (b).

Solution

1. Apply the power law of logarithms.

   Using the identity ${\log }_a(m^n)=n{\log }_{a}m$:

   $$\begin{array}{cc}{\log }_2(2^3)& =3{\log }_{2}2\end{array}$$

2. Apply the base identity ${\log }_{a}a=1$.

   Since ${\log }_{2}2=1$:

   $$\begin{array}{cc}& =3(1)\\ & =3\end{array}$$

3. Identify the correct option.

   This corresponds to option (d).

Solution

1. Recognize the common logarithm base.

   $\log 100$ represents ${\log }_{10}100$.

2. Express $100$ as a power of $10$.

   Since $100={10}^2$:

   $$\begin{array}{cc}{\log }_{10}100& ={\log }_{10}({10}^2)\end{array}$$

3. Apply the power law and base identity.

   $$\begin{array}{cc}& =2{\log }_{10}10\\ & =2(1)\\ & =2\end{array}$$

4. Identify the correct option.

   This matches option (a).

Solution

1. Recall the Product Law of Logarithms.

   The product law of logarithms states that:

   $${\log }_b(m\times n)={\log }_{b}m+{\log }_{b}n$$

2. Decompose the argument $200$.

   Write $200$ in terms of the given value $2$ and a power of $10$:

   $$200=2\times 100=2\times {10}^2$$

3. Apply logarithm laws step by step.

   $$\begin{array}{cc}\log 200& =\log (2\times 100)\\ & =\log 2+\log 100\\ & =\log 2+{\log }_{10}({10}^2)\\ & =\log 2+2{\log }_{10}(10)\\ & =\log 2+2(1)\\ & =\log 2+2\end{array}$$

4. Substitute the given numerical value $\log 2=0.3010$.

   $$\begin{array}{cc}\log 200& =0.3010+2\\ & =2.3010\end{array}$$

5. Identify the correct option.

   Comparing with the options:

   • (a) $1.3010$
   • (b) $0.6010$
   • (c) $2.3010$ $✓$
   • (d) $2.6010$

   Hence, the correct option is (c).

Solution

1. Recall the domain of logarithmic functions.

   For any real base $b>0$ ($b\ne 1$), the logarithm ${\log }_b(x)$ is defined strictly for $x>0$.

2. Analyze $\log (0)$.

   There is no real exponent $y$ such that ${10}^y=0$, because ${10}^y>0$ for all real numbers $y$. As $x\to 0^{+}$, ${\log }_{10}(x)\to -\infty$.

3. Conclude.

   Hence, $\log (0)$ is undefined in the real number system, which corresponds to option (d).

Solution

1. Express $\mathrm{10,000}$ as a power of $10$.

   Since $\mathrm{10,000}={10}^4$:

   $$\begin{array}{cc}\log \mathrm{10,000}& ={\log }_{10}({10}^4)\end{array}$$

2. Apply the power law of logarithms.

   $$\begin{array}{cc}& =4{\log }_{10}10\\ & =4(1)\\ & =4\end{array}$$

3. Identify the correct option.

   This matches option (c).

Solution

1. Recall the product law of logarithms.

   According to the first law of logarithms:

   $$\begin{array}{cc}{\log }_{a}m+{\log }_{a}n& ={\log }_a(m\times n)\end{array}$$

2. Apply the law to the given expression.

   $$\begin{array}{cc}\log 5+\log 3& =\log (5\times 3)\\ & =\log 15\end{array}$$

3. Identify the correct option.

   This corresponds to option (d).

Solution

1. Recall the definition of logarithm.

   The exponential relationship $a^x=y$ is equivalent to the logarithmic form:

   $$\begin{array}{cc}{\log }_{a}y& =x\end{array}$$

2. Identify base, exponent, and result.

   In the equation $3^4=81$:

   • Base: $a=3$
   • Exponent: $x=4$
   • Result: $y=81$

3. Write in logarithmic form.

   $$\begin{array}{cc}{\log }_{3}81& =4\end{array}$$

4. Identify the correct option.

   This corresponds to option (c).

Solution

1. Understand scientific notation format.

   Scientific notation has the form $A\times {10}^n$, where $1\le A<10$ and $n\in \mathbb{Z}$.

2. Place the decimal point after the first non-zero digit.

   The first non-zero digit in $0.000567$ is $5$. Placing the decimal point after $5$ gives:

   $$\begin{array}{cc}A& =5.67\end{array}$$

3. Count the number of places shifted.

   The decimal point is moved $4$ places to the right. Since $0.000567<1$, the power of $10$ is negative:

   $$\begin{array}{cc}0.000567& =5.67\times {10}^{-4}\end{array}$$

Solution

1. Locate the current decimal position.

   The number is $734$, which can be written as $734.0$.

2. Place the decimal point after the first non-zero digit.

   The first non-zero digit is $7$. Placing the decimal point after $7$ gives $7.34$.

3. Determine the power of 10.

   We move the decimal point $2$ places to the left. Since $734>10$, the power of $10$ is positive:

   $$\begin{array}{cc}734& =7.34\times {10}^2\end{array}$$

Solution

1. Express $0.33$ in standard scientific notation.

   Moving the decimal point $1$ place to the right gives:

   $$\begin{array}{cc}0.33& =3.3\times {10}^{-1}\end{array}$$

2. Substitute back into the expression.

   $$\begin{array}{cc}0.33\times {10}^3& =(3.3\times {10}^{-1})\times {10}^3\end{array}$$

3. Apply the product rule of exponents $a^m\cdot a^n=a^{m+n}$.

   $$\begin{array}{cc}& =3.3\times {10}^{-1+3}\\ & =3.3\times {10}^2\end{array}$$

Solution

1. Identify the power of 10.

   The given expression is $2.6\times {10}^3$. The positive power $3$ indicates multiplication by $\mathrm{1,000}$.

2. Move the decimal point $3$ places to the right.

   $$\begin{array}{cc}2.6\times {10}^3& =2.6\times \mathrm{1,000}\\ & =2600\end{array}$$

Solution

1. Identify the power of 10.

   The given expression is $8.794\times {10}^{-4}$. The negative exponent $-4$ indicates division by ${10}^4=\mathrm{10,000}$.

2. Shift the decimal point $4$ places to the left.

   $$\begin{array}{cc}8.794\times {10}^{-4}& =\frac{8.794}{\mathrm{10,000}}\\ & =0.0008794\end{array}$$

Solution

1. Identify the power of 10.

   The given number is $6\times {10}^{-6}=6.0\times {10}^{-6}$.

2. Shift the decimal point $6$ places to the left.

   Dividing by ${10}^6=\mathrm{1,000,000}$ places $5$ zeros between the decimal point and the digit $6$:

   $$\begin{array}{cc}6\times {10}^{-6}& =\frac{6}{\mathrm{1,000,000}}\\ & =0.000006\end{array}$$

Solution

1. Recall the relationship between exponential and logarithmic form.

   An exponential statement $a^x=y$ is converted to logarithmic form as:

   $$\begin{array}{cc}{\log }_{a}y& =x\end{array}$$

2. Identify base, exponent, and value.

   From $3^7=2187$:

   • Base: $a=3$
   • Exponent: $x=7$
   • Value: $y=2187$

3. Write in logarithmic form.

   $$\begin{array}{cc}{\log }_{3}2187& =7\end{array}$$

Solution

1. Recall the general definition of logarithm.

   If $a^x=y$ where $a>0,a\ne 1$ and $y>0$, then:

   $$\begin{array}{cc}{\log }_{a}y& =x\end{array}$$

2. Map variables to the definition.

   In the equation $a^b=c$:

   • Base: $a$
   • Exponent: $b$
   • Value: $c$

3. Express in logarithmic form.

   $$\begin{array}{cc}{\log }_{a}c& =b\end{array}$$

Solution

1. Recall the logarithmic equivalence.

   $$\begin{array}{cc}{a}^x=y& ⟺{\log }_{a}y=x\end{array}$$

2. Identify the components.

   From $(12{)}^2=144$:

   • Base: $a=12$
   • Exponent: $x=2$
   • Value: $y=144$

3. Write in logarithmic form.

   $$\begin{array}{cc}{\log }_{12}144& =2\end{array}$$

Solution

1. Recall the conversion formula.

   A logarithmic equation ${\log }_{a}y=x$ converts to exponential form as:

   $$\begin{array}{cc}{a}^x& =y\end{array}$$

2. Identify base, exponent, and value.

   From ${\log }_{4}8=x$:

   • Base: $a=4$
   • Exponent: $x$
   • Value: $y=8$

3. Write in exponential form.

   $$\begin{array}{cc}{4}^x& =8\end{array}$$

Solution

1. Recall the conversion formula.

   $$\begin{array}{cc}{\log }_{a}y=x& ⟺a^x=y\end{array}$$

2. Identify base, exponent, and value.

   From ${\log }_{9}729=3$:

   • Base: $a=9$
   • Exponent: $x=3$
   • Value: $y=729$

3. Write in exponential form.

   $$\begin{array}{cc}{9}^3& =729\end{array}$$

Solution

1. Recall the conversion formula.

   $$\begin{array}{cc}{\log }_{a}y=x& ⟺a^x=y\end{array}$$

2. Identify base, exponent, and value.

   From ${\log }_{4}1024=5$:

   • Base: $a=4$
   • Exponent: $x=5$
   • Value: $y=1024$

3. Write in exponential form.

   $$\begin{array}{cc}{4}^5& =1024\end{array}$$

Solution

1. Convert the logarithmic equation to exponential form.

   Recall that ${\log }_{a}y=x⟺y=a^x$:

   $$\begin{array}{cc}{\log }_{9}x& =0.5\\ x& =9^{0.5}\end{array}$$

2. Express the decimal power as a fraction.

   Since $0.5=\frac{1}{2}$:

   $$\begin{array}{cc}x& =9^{\frac{1}{2}}\\ & =\sqrt{9}\\ & =\sqrt{3^2}\end{array}$$

3. Conclude the value of $x$.

   $$\begin{array}{cc}x& =3\end{array}$$

Solution

1. Express both sides in terms of common base $3$.

   Since $\frac{1}{9}=\frac{1}{3^2}=3^{-2}$ and $27=3^3$:

   $$\begin{array}{cc}{\left(\frac{1}{9}\right)}^{3x}& =27\\ {\left(\frac{1}{3^2}\right)}^{3x}& =3^3\\ (3^{-2}{)}^{3x}& =3^3\end{array}$$

2. Apply power of a power rule $(a^m{)}^n=a^{mn}$.

   $$\begin{array}{cc}{3}^{(-2)(3x)}& =3^3\\ 3^{-6x}& =3^3\end{array}$$

3. Equate exponents with matching bases.

   Since bases are equal:

   $$\begin{array}{cc}-6x& =3\\ x& =-\frac{3}{6}\\ x& =-\frac{1}{2}\end{array}$$

Solution

1. Express both sides in terms of common base $2$.

   Since $\frac{1}{32}=\frac{1}{2^5}=2^{-5}$ and $64=2^6$:

   $$\begin{array}{cc}{\left(\frac{1}{32}\right)}^{2x}& =64\\ {\left(\frac{1}{2^5}\right)}^{2x}& =2^6\\ (2^{-5}{)}^{2x}& =2^6\end{array}$$

2. Multiply exponents using $(a^m{)}^n=a^{mn}$.

   $$\begin{array}{cc}{2}^{(-5)(2x)}& =2^6\\ 2^{-10x}& =2^6\end{array}$$

3. Equate exponents and solve for $x$.

   Since the bases are identical:

   $$\begin{array}{cc}-10x& =6\\ x& =-\frac{6}{10}\\ x& =-\frac{3}{5}\end{array}$$

Solution

1. Apply the power law of logarithms $n\log a=\log (a^n)$.

   $$\begin{array}{cc}7\log x-3\log y^2& =\log (x^7)-\log ((y^2{)}^3)\end{array}$$

2. Simplify powers in the second argument.

   $$\begin{array}{cc}& =\log (x^7)-\log (y^6)\end{array}$$

3. Apply the quotient law $\log m-\log n=\log \left(\frac{m}{n}\right)$.

   $$\begin{array}{cc}& =\log \left(\frac{x^7}{y^6}\right)\end{array}$$

Solution

1. Move coefficient into exponent using the power law.

   $$\begin{array}{cc}3\log 4-\log 32& =\log (4^3)-\log 32\end{array}$$

2. Evaluate the power $4^3=64$.

   $$\begin{array}{cc}& =\log 64-\log 32\end{array}$$

3. Apply quotient law and simplify.

   $$\begin{array}{cc}& =\log \left(\frac{64}{32}\right)\\ & =\log 2\end{array}$$

Solution

1. Apply the product law inside parentheses.

   $$\begin{array}{cc}\frac{1}{3}({\log }_{5}8+{\log }_{5}27)-{\log }_{5}3& =\frac{1}{3}{\log }_5(8\times 27)-{\log }_{5}3\\ & =\frac{1}{3}{\log }_5(216)-{\log }_{5}3\end{array}$$

2. Apply the power law to the first term.

   Since $216=6^3$:

   $$\begin{array}{cc}& ={\log }_5\left({216}^{\frac{1}{3}}\right)-{\log }_{5}3\\ & ={\log }_5((6^3{)}^{\frac{1}{3}})-{\log }_{5}3\\ & ={\log }_{5}6-{\log }_{5}3\end{array}$$

3. Apply the quotient law and simplify.

   $$\begin{array}{cc}& ={\log }_5\left(\frac{6}{3}\right)\\ & ={\log }_{5}2\end{array}$$

Solution

1. Apply the product law of logarithms.

   $$\begin{array}{cc}\log (xy{z}^6)& =\log x+\log y+\log (z^6)\end{array}$$

2. Apply the power law $\log (z^n)=n\log z$.

   $$\begin{array}{cc}& =\log x+\log y+6\log z\end{array}$$

Solution

1. Convert radical expression to rational exponential form.

   $$\begin{array}{cc}{\log }_3\sqrt[6]{m^5{n}^3}& ={\log }_3((m^5{n}^3{)}^{\frac{1}{6}})\end{array}$$

2. Apply power law ${\log }_a(u^k)=k{\log }_{a}u$.

   $$\begin{array}{cc}& =\frac{1}{6}{\log }_3(m^5{n}^3)\end{array}$$

3. Apply product law and distribute the exponent.

   $$\begin{array}{cc}& =\frac{1}{6}[{\log }_3(m^5)+{\log }_3(n^3)]\\ & =\frac{1}{6}[5{\log }_{3}m+3{\log }_{3}n]\\ & =\frac{5}{6}{\log }_{3}m+\frac{3}{6}{\log }_{3}n\\ & =\frac{5}{6}{\log }_{3}m+\frac{1}{2}{\log }_{3}n\end{array}$$

Solution

1. Express square root as rational power $\frac{1}{2}$ and factorize $8=2^3$.

   $$\begin{array}{cc}\log \sqrt{8x^3}& =\log ((8x^3{)}^{\frac{1}{2}})\\ & =\log ((2^3\cdot x^3{)}^{\frac{1}{2}})\end{array}$$

2. Apply power law of logarithms.

   $$\begin{array}{cc}& =\frac{1}{2}\log (2^3\cdot x^3)\end{array}$$

3. Apply product and power laws to expand each factor.

   $$\begin{array}{cc}& =\frac{1}{2}[\log (2^3)+\log (x^3)]\\ & =\frac{1}{2}[3\log 2+3\log x]\\ & =\frac{3}{2}(\log 2+\log x)\\ & =\frac{3}{2}\log 2+\frac{3}{2}\log x\end{array}$$

Solution

1. Let the given expression be $x$ and apply logarithm on both sides.

   $$\begin{array}{cc}x& =\sqrt[3]{68.24}=(68.24{)}^{\frac{1}{3}}\\ \log x& =\log ((68.24{)}^{\frac{1}{3}})\\ \log x& =\frac{1}{3}\log (68.24)\end{array}$$

2. Find $\log (68.24)$ using logarithm table.

   • Characteristic: Since there are $2$ digits before decimal point in $68.24$, $\text{Characteristic}=2-1=1$.
   • Mantissa: In the log table, row $68$ under column $2$ gives $8338$. Mean difference under column $4$ is $2$.

   $$\text{Mantissa}=0.8338+0.0002=0.8340$$ $$\begin{array}{cc}\log (68.24)& =1.8340\end{array}$$

3. Divide by $3$.

   $$\begin{array}{cc}\log x& =\frac{1.8340}{3}\\ & =0.6113\end{array}$$

4. Take antilogarithm to find $x$.

   • Characteristic $=0⟹$ number of integral digits $=0+1=1$.
   • Antilog table for $.6113$: Row $.61$ under column $1$ is $4083$, mean difference for $3$ is $3$:

   $$4083+3=4086$$ $$\begin{array}{cc}x& =\text{antilog}(0.6113)\\ x& =4.086\end{array}$$

Solution

1. Let $x=319.8\times 3.543$ and apply logarithm on both sides.

   $$\begin{array}{cc}\log x& =\log (319.8\times 3.543)\\ & =\log (319.8)+\log (3.543)\end{array}$$

2. Find logarithms of the factors using log table.

   • For $319.8$:
     • Characteristic $=3-1=2$.
     • Mantissa: Row $31$, column $9$ is $5038$; mean difference for $8$ is $11$: $$\text{Mantissa}=0.5038+0.0011=0.5049⟹\log (319.8)=2.5049$$
   • For $3.543$:
     • Characteristic $=1-1=0$.
     • Mantissa: Row $35$, column $4$ is $5490$; mean difference for $3$ is $4$: $$\text{Mantissa}=0.5490+0.0004=0.5494⟹\log (3.543)=0.5494$$

3. Add the logarithms.

   $$\begin{array}{cc}\log x& =2.5049+0.5494\\ & =3.0543\end{array}$$

4. Find antilogarithm of $3.0543$.

   • Characteristic $=3⟹$ number of integral digits $=3+1=4$.
   • Antilog table for $.0543$: Row $.05$, column $4$ is $1132$; mean difference for $3$ is $1$: $$1132+1=1133$$

   $$\begin{array}{cc}x& =\text{antilog}(3.0543)\\ x& =1133\end{array}$$

Solution

1. Let $x$ be the expression and apply logarithm.

   $$\begin{array}{cc}x& =\frac{36.12\times 750.9}{113.2\times 9.98}\\ \log x& =\log \left(\frac{36.12\times 750.9}{113.2\times 9.98}\right)\\ & =[\log (36.12)+\log (750.9)]-[\log (113.2)+\log (9.98)]\end{array}$$

2. Find individual logarithms using tables.

   • $\log (36.12)=1.5577$
   • $\log (750.9)=2.8756$
   • $\log (113.2)=2.0538$
   • $\log (9.98)=0.9991$

3. Simplify numerator and denominator sums.

   $$\begin{array}{cc}\log x& =(1.5577+2.8756)-(2.0538+0.9991)\\ & =4.4333-3.0529\\ & =1.3804\end{array}$$

4. Find antilogarithm of $1.3804$.

   • Characteristic $=1⟹$ number of integral digits $=1+1=2$.
   • Antilog table for $.3804$: Row $.38$, column $0$ is $2399$; mean difference for $4$ is $2$: $$2399+2=2401$$

   $$\begin{array}{cc}x& =\text{antilog}(1.3804)\\ x& =24.01\end{array}$$