Answer :
Let's simplify each part of the question step-by-step:
1.1 Simplify the following WITHOUT using a calculator:
1.1.1 [tex]\sqrt{49x^4}[/tex]
The square root of a product can be broken down into the product of square roots:
[tex]\sqrt{49x^4} = \sqrt{49} \cdot \sqrt{x^4}[/tex]
Since [tex]\sqrt{49} = 7[/tex] and [tex]\sqrt{x^4} = x^2[/tex], we get:
[tex]7x^2[/tex]
1.1.2 [tex]\frac{\sqrt[3]{49^4}}{\sqrt{16x^2 + 9x^2}}[/tex]
First, simplify the expression in the denominator:
[tex]16x^2 + 9x^2 = 25x^2[/tex]
Thus, [tex]\sqrt{25x^2} = 5x[/tex].
In the numerator, [tex]\sqrt[3]{49^4} = (49^{\frac{4}{3}})[/tex] can be rewritten as a power:
[tex](49^{\frac{4}{3}}) = (7^2)^{\frac{4}{3}} = 7^{\frac{8}{3}}[/tex]
Thus, [tex]\frac{7^{\frac{8}{3}}}{5x}[/tex].
1.1.3 [tex]3^{x+1} + 3^x[/tex]
Using exponent rules, we know [tex]3^{x+1} = 3 \cdot 3^x[/tex], so the expression becomes:
[tex]3 \cdot 3^x + 3^x = 4 \cdot 3^x[/tex]
1.1.4 [tex]\log 40 + \log 5 - \log 2[/tex]
Using log properties, we combine the logs:
[tex]\log(40 \times 5) - \log 2[/tex]
This simplifies to:
[tex]\log 200 - \log 2 = \log \left(\frac{200}{2}\right) = \log 100[/tex]
Since [tex]\log 100 = 2[/tex], the answer is 2.
1.1.5 [tex]\log_5\left(\frac{1}{5}\right) + \log_5 30 - \log_5 6[/tex]
Using log properties, we have:
[tex]\log_5\left(\frac{1}{5}\right) = -1[/tex] (because [tex]\log_5 5 = 1[/tex] and [tex]\log_5 1 = 0[/tex])
Combine the logs using the property:
[tex]-1 + \log_5\left(\frac{30}{6}\right) = -1 + \log_5 5 = -1 + 1 = 0[/tex]
1.2 Show that:
[tex]\log 6 + 2 \log 20 - \log 3 - 3 \log 2 = 2[/tex]
Using properties of logarithms:
[tex]\log 6 + 2 \log 20 - \log 3 - 3 \log 2[/tex]
[tex]= \log 6 + \log 20^2 - \log 3 - \log 8[/tex]
[tex]= \log \left(\frac{6 \times 400}{3 \times 8}\right)[/tex]
[tex]= \log \left(\frac{2400}{24}\right) = \log 100[/tex]
Since [tex]\log 100 = 2[/tex], the equation is shown.
1.3 Simplify the following surd numbers:
[tex](2 + 3\sqrt{5})(2 - 3\sqrt{5})[/tex]
This is a difference of squares:
[tex](2 + 3\sqrt{5})(2 - 3\sqrt{5}) = 2^2 - (3\sqrt{5})^2[/tex]
[tex]= 4 - 45 = -41[/tex]
2.1 Solve for [tex]x[/tex]:
2.1.1 [tex]2^{-1} = 64[/tex]
[tex]2^{-x} = 64[/tex] implies [tex]\frac{1}{2^x} = 64[/tex], so [tex]2^x = \frac{1}{64}[/tex].
Thus, [tex]x = -6[/tex].
2.1.2 [tex]5x^2 - 4 - 1 = 0[/tex]
Simplify equation:
[tex]5x^2 - 5 = 0[/tex]
[tex]5x^2 = 5[/tex]
Divide by 5:
[tex]x^2 = 1[/tex]
[tex]x = \pm 1[/tex]
2.1.3 [tex]x^2 - 4x - 3 = 0[/tex], Round off to ONE decimal place.
Use quadratic formula: [tex]x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}[/tex], where [tex]a = 1, b = -4, c = -3[/tex]:
[tex]x = \frac{4 \pm \sqrt{16 + 12}}{2}[/tex]
[tex]x = \frac{4 \pm \sqrt{28}}{2}[/tex]
[tex]x = 2 \pm \sqrt{7}[/tex]
Rounded to one decimal place:
[tex]x \approx 4.6[/tex] or [tex]x \approx -0.6[/tex]
2.1.4 [tex]x^2 - 5x \leq -6[/tex], represent your solution on a number line.
Rearrange:
[tex]x^2 - 5x + 6 \leq 0[/tex]
Factorize:
[tex](x - 3)(x - 2) \leq 0[/tex]
The critical points are [tex]2[/tex] and [tex]3[/tex]. Test intervals:
- For [tex]x < 2[/tex], [tex](-)\cdot (-) > 0[/tex]
- For [tex]2 < x < 3[/tex], [tex](-)\cdot (+) < 0[/tex]
- For [tex]x > 3[/tex], [tex](+)\cdot (+) > 0[/tex]
Solution in interval notation: [tex]2 \leq x \leq 3[/tex]
Number line:
-----------------|===========|----------------2 3
2.1.5 [tex]\log_2 32 = x[/tex]
Since [tex]32 = 2^5[/tex], [tex]x = 5[/tex].
2.1.6 [tex]2 \log x + 2 = \log 100[/tex]
Since [tex]\log 100 = 2[/tex], we have:
[tex]2 \log x + 2 = 2[/tex]
[tex]2 \log x = 0[/tex]
[tex]\log x = 0[/tex]
[tex]x = 10^0 = 1[/tex]
I hope this helps! Please let me know if you need further clarification on any part.