a) Using log table, evaluate (3.678)^4 [3]42.75 [3] b) Evaluate : x 3 x^3 27x^2 5x + 6 [3]

a) Evaluate $\frac{(3.678{)}^4}{\sqrt[3]{42.75}}$ using log-table

Step 1: Break down the expression

The given expression is: $$\frac{(3.678{)}^4}{\sqrt[3]{42.75}}=\frac{(3.678{)}^4}{(42.75{)}^{1/3}}$$

Step 2: Apply logarithm properties

We use the property of logarithms: $$\log \left(\frac{a}{b}\right)=\log a-\log b$$ and $$\log (a^m)=m\log a$$ Thus, $$\log \left(\frac{(3.678{)}^4}{(42.75{)}^{1/3}}\right)=4\log (3.678)-\frac{1}{3}\log (42.75)$$

Step 3: Find $\log (3.678)$ using log-table

From the log-table (assuming base 10):

• $\log 3.6=0.5563$
• The difference for 0.07 (since 3.678 - 3.6 = 0.078, but we approximate to 0.07 for simplicity):
  • Mean difference for 0.07 in the table is approximately 0.0031.
• Thus, $$\log (3.678)\approx 0.5563+0.0031=0.5594$$

Step 4: Calculate $4\log (3.678)$

$$4\log (3.678)=4\times 0.5594=2.2376$$

Step 5: Find $\log (42.75)$ using log-table

From the log-table:

• $\log 42=1.6232$
• $\log 43=1.6335$
• Difference between $\log 42$ and $\log 43$ is $0.0103$.
• For 42.75, we interpolate:
  • Fractional part: $0.75\times 0.0103=0.0077$
  • Thus, $$\log (42.75)\approx 1.6232+0.0077=1.6309$$

Step 6: Calculate $\frac{1}{3}\log (42.75)$

$$\frac{1}{3}\log (42.75)=\frac{1.6309}{3}=0.5436$$

Step 7: Compute the logarithm of the expression

$$\log \left(\frac{(3.678{)}^4}{\sqrt[3]{42.75}}\right)=2.2376-0.5436=1.6940$$

Step 8: Find the antilogarithm

Using the antilog-table:

• The characteristic is 1, so the number is between 10 and 100.
• The mantissa is 0.6940.
• From the antilog-table, 0.6940 corresponds to approximately 4.94.
• Thus, $$\frac{(3.678{)}^4}{\sqrt[3]{42.75}}\approx 10^{1.6940}=10^{1+0.6940}=10^1\times 10^{0.6940}=10\times 4.94=49.4$$

Final Answer:

$$\boxed{49.4}$$

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b) Evaluate ${\lim }_{x\to 3}\frac{x^3-27}{x^2-5x+6}$

Step 1: Direct substitution

First, substitute $x=3$ directly into the expression: $$\frac{3^3-27}{3^2-5\times 3+6}=\frac{27-27}{9-15+6}=\frac{0}{0}$$ This is an indeterminate form, so we must simplify the expression.

Step 2: Factorize the numerator and denominator

• Numerator: $x^3-27$ is a difference of cubes, which factors as: $$x^3-27=(x-3)(x^2+3x+9)$$

• Denominator: $x^2-5x+6$ factors as: $$x^2-5x+6=(x-2)(x-3)$$

Step 3: Simplify the expression

$$\frac{x^3-27}{x^2-5x+6}=\frac{(x-3)(x^2+3x+9)}{(x-2)(x-3)}$$ Cancel the common factor $(x-3)$ (valid since $x\ne 3$): $$\frac{x^2+3x+9}{x-2}$$

Step 4: Evaluate the limit after simplification

Now, substitute $x=3$ into the simplified expression: $$\frac{3^2+3\times 3+9}{3-2}=\frac{9+9+9}{1}=\frac{27}{1}=27$$

Final Answer:

$$\boxed{27}$$