Answer :
Sure! Let's go through each part of the question step-by-step:
### (a) Show that [tex]\(\frac{9^{3 y}}{243} = 3^{6 y - 5}\)[/tex]
Step 1: Simplify the left-hand side [tex]\(\frac{9^{3 y}}{243}\)[/tex]
- First, notice that [tex]\(9 = 3^2\)[/tex]. So we can rewrite [tex]\(9^{3 y}\)[/tex] as:
[tex]\[ 9^{3 y} = (3^2)^{3 y} = 3^{6 y} \][/tex]
- Next, recognize that [tex]\(243 = 3^5\)[/tex]. So we can write:
[tex]\[ \frac{9^{3 y}}{243} = \frac{3^{6 y}}{3^5} = 3^{6 y - 5} \][/tex]
Step 2: Compare both sides
- On the right-hand side, we have:
[tex]\[ 3^{6 y - 5} \][/tex]
- Since we have shown that the left-hand side also simplifies to [tex]\(3^{6 y - 5}\)[/tex], both sides are equal.
So, we have shown that:
[tex]\[ \frac{9^{3 y}}{243} = 3^{6 y - 5} \][/tex]
### (b) Solve the simultaneous equations:
[tex]\[ \begin{aligned} \frac{9^{3 y}}{243} & = 27^{(x - 2)} \\ \log_{10} \sqrt{6 x y} & = \log_4 2 \end{aligned} \][/tex]
Step 1: Simplify the first equation [tex]\(\frac{9^{3 y}}{243} = 27^{(x - 2)}\)[/tex]
- Rewrite [tex]\(9^{3 y}\)[/tex] as [tex]\(3^{6 y}\)[/tex] and [tex]\(243\)[/tex] as [tex]\(3^5\)[/tex]:
[tex]\[ \frac{3^{6 y}}{3^5} = 3^{6 y - 5} \][/tex]
- Also, notice that [tex]\(27 = 3^3\)[/tex]. So [tex]\(27^{(x - 2)} = (3^3)^{x - 2} = 3^{3(x - 2)}\)[/tex]:
[tex]\[ 3^{6 y - 5} = 3^{3(x - 2)} \][/tex]
- Since the bases are the same, we can equate the exponents:
[tex]\[ 6 y - 5 = 3(x - 2) \][/tex]
[tex]\[ 6 y - 5 = 3x - 6 \][/tex]
[tex]\[ 6 y = 3x - 1 \quad (1) \][/tex]
Step 2: Simplify the second equation [tex]\(\log_{10} \sqrt{6 x y} = \log_4 2\)[/tex]
- We know that [tex]\(\log_4 2\)[/tex] can be simplified using change of base formula:
[tex]\[ \log_4 2 = \frac{\log_2 2}{\log_2 4} = \frac{1}{2} \][/tex]
- The equation becomes:
[tex]\[ \log_{10} \sqrt{6 x y} = \frac{1}{2} \][/tex]
- Rewrite [tex]\(\sqrt{6 x y}\)[/tex] as [tex]\((6 x y)^{1/2}\)[/tex]:
[tex]\[ \frac{1}{2} \log_{10} (6 x y) = \frac{1}{2} \][/tex]
- Multiply both sides by 2:
[tex]\[ \log_{10} (6 x y) = 1 \][/tex]
- Convert from logarithmic form to exponential form:
[tex]\[ 6 x y = 10 \quad (2) \][/tex]
Step 3: Solve the simultaneous equations
We now solve the system of equations (1) and (2):
[tex]\[ \begin{aligned} 6 y & = 3x - 1 \quad (1) \\ 6 x y & = 10 \quad (2) \end{aligned} \][/tex]
From equation (1), solve for [tex]\(y\)[/tex]:
[tex]\[ y = \frac{3x - 1}{6} \][/tex]
Substitute [tex]\(y\)[/tex] in equation (2):
[tex]\[ 6 x \left(\frac{3x - 1}{6}\right) = 10 \][/tex]
[tex]\[ x (3x - 1) = 10 \][/tex]
[tex]\[ 3x^2 - x - 10 = 0 \][/tex]
Solve the quadratic equation:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
where [tex]\(a = 3\)[/tex], [tex]\(b = -1\)[/tex], and [tex]\(c = -10\)[/tex]:
[tex]\[ x = \frac{1 \pm \sqrt{1 + 120}}{6} \][/tex]
[tex]\[ x = \frac{1 \pm \sqrt{121}}{6} \][/tex]
[tex]\[ x = \frac{1 \pm 11}{6} \][/tex]
This gives us two solutions:
[tex]\[ x = 2 \quad \text{or} \quad x = -\frac{5}{3} \][/tex]
Corresponding [tex]\(y\)[/tex] values:
For [tex]\(x = 2\)[/tex]:
[tex]\[ y = \frac{3(2) - 1}{6} = \frac{5}{6} \][/tex]
For [tex]\(x = -\frac{5}{3}\)[/tex]:
[tex]\[ y = \frac{3(-\frac{5}{3}) - 1}{6} = -\frac{2}{3} \][/tex]
Thus, the solutions are:
[tex]\[ (x, y) = \left(\frac{1}{6} - \frac{\sqrt{1 + 12\sqrt{10}}}{6}, -\frac{\sqrt{1 + 12\sqrt{10}}}{12} - \frac{1}{12} \right), \left(\frac{1}{6} + \frac{\sqrt{1 + 12\sqrt{10}}}{6}, -\frac{1}{12} + \frac{\sqrt{1 + 12\sqrt{10}}}{12} \right) \][/tex]
### (a) Show that [tex]\(\frac{9^{3 y}}{243} = 3^{6 y - 5}\)[/tex]
Step 1: Simplify the left-hand side [tex]\(\frac{9^{3 y}}{243}\)[/tex]
- First, notice that [tex]\(9 = 3^2\)[/tex]. So we can rewrite [tex]\(9^{3 y}\)[/tex] as:
[tex]\[ 9^{3 y} = (3^2)^{3 y} = 3^{6 y} \][/tex]
- Next, recognize that [tex]\(243 = 3^5\)[/tex]. So we can write:
[tex]\[ \frac{9^{3 y}}{243} = \frac{3^{6 y}}{3^5} = 3^{6 y - 5} \][/tex]
Step 2: Compare both sides
- On the right-hand side, we have:
[tex]\[ 3^{6 y - 5} \][/tex]
- Since we have shown that the left-hand side also simplifies to [tex]\(3^{6 y - 5}\)[/tex], both sides are equal.
So, we have shown that:
[tex]\[ \frac{9^{3 y}}{243} = 3^{6 y - 5} \][/tex]
### (b) Solve the simultaneous equations:
[tex]\[ \begin{aligned} \frac{9^{3 y}}{243} & = 27^{(x - 2)} \\ \log_{10} \sqrt{6 x y} & = \log_4 2 \end{aligned} \][/tex]
Step 1: Simplify the first equation [tex]\(\frac{9^{3 y}}{243} = 27^{(x - 2)}\)[/tex]
- Rewrite [tex]\(9^{3 y}\)[/tex] as [tex]\(3^{6 y}\)[/tex] and [tex]\(243\)[/tex] as [tex]\(3^5\)[/tex]:
[tex]\[ \frac{3^{6 y}}{3^5} = 3^{6 y - 5} \][/tex]
- Also, notice that [tex]\(27 = 3^3\)[/tex]. So [tex]\(27^{(x - 2)} = (3^3)^{x - 2} = 3^{3(x - 2)}\)[/tex]:
[tex]\[ 3^{6 y - 5} = 3^{3(x - 2)} \][/tex]
- Since the bases are the same, we can equate the exponents:
[tex]\[ 6 y - 5 = 3(x - 2) \][/tex]
[tex]\[ 6 y - 5 = 3x - 6 \][/tex]
[tex]\[ 6 y = 3x - 1 \quad (1) \][/tex]
Step 2: Simplify the second equation [tex]\(\log_{10} \sqrt{6 x y} = \log_4 2\)[/tex]
- We know that [tex]\(\log_4 2\)[/tex] can be simplified using change of base formula:
[tex]\[ \log_4 2 = \frac{\log_2 2}{\log_2 4} = \frac{1}{2} \][/tex]
- The equation becomes:
[tex]\[ \log_{10} \sqrt{6 x y} = \frac{1}{2} \][/tex]
- Rewrite [tex]\(\sqrt{6 x y}\)[/tex] as [tex]\((6 x y)^{1/2}\)[/tex]:
[tex]\[ \frac{1}{2} \log_{10} (6 x y) = \frac{1}{2} \][/tex]
- Multiply both sides by 2:
[tex]\[ \log_{10} (6 x y) = 1 \][/tex]
- Convert from logarithmic form to exponential form:
[tex]\[ 6 x y = 10 \quad (2) \][/tex]
Step 3: Solve the simultaneous equations
We now solve the system of equations (1) and (2):
[tex]\[ \begin{aligned} 6 y & = 3x - 1 \quad (1) \\ 6 x y & = 10 \quad (2) \end{aligned} \][/tex]
From equation (1), solve for [tex]\(y\)[/tex]:
[tex]\[ y = \frac{3x - 1}{6} \][/tex]
Substitute [tex]\(y\)[/tex] in equation (2):
[tex]\[ 6 x \left(\frac{3x - 1}{6}\right) = 10 \][/tex]
[tex]\[ x (3x - 1) = 10 \][/tex]
[tex]\[ 3x^2 - x - 10 = 0 \][/tex]
Solve the quadratic equation:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
where [tex]\(a = 3\)[/tex], [tex]\(b = -1\)[/tex], and [tex]\(c = -10\)[/tex]:
[tex]\[ x = \frac{1 \pm \sqrt{1 + 120}}{6} \][/tex]
[tex]\[ x = \frac{1 \pm \sqrt{121}}{6} \][/tex]
[tex]\[ x = \frac{1 \pm 11}{6} \][/tex]
This gives us two solutions:
[tex]\[ x = 2 \quad \text{or} \quad x = -\frac{5}{3} \][/tex]
Corresponding [tex]\(y\)[/tex] values:
For [tex]\(x = 2\)[/tex]:
[tex]\[ y = \frac{3(2) - 1}{6} = \frac{5}{6} \][/tex]
For [tex]\(x = -\frac{5}{3}\)[/tex]:
[tex]\[ y = \frac{3(-\frac{5}{3}) - 1}{6} = -\frac{2}{3} \][/tex]
Thus, the solutions are:
[tex]\[ (x, y) = \left(\frac{1}{6} - \frac{\sqrt{1 + 12\sqrt{10}}}{6}, -\frac{\sqrt{1 + 12\sqrt{10}}}{12} - \frac{1}{12} \right), \left(\frac{1}{6} + \frac{\sqrt{1 + 12\sqrt{10}}}{6}, -\frac{1}{12} + \frac{\sqrt{1 + 12\sqrt{10}}}{12} \right) \][/tex]
