Question 10
(a) Show that $\frac{9^{3 y}}{243}=3^{(6 y-5)}$
(4)
(b) Solve the simultaneous equations
$$ \begin{aligned} \frac{9^{3 y}}{243} & =27^{(x-2)} \\ \log _{10} \sqrt{6 x y} & =\log _4 2 \end{aligned} $$
(9)
Solution
(a) Show that $\frac{9^{3y}}{243} = 3^{6y-5}$
Step 1: Rewrite the base $9$ and $243$ in terms of $3$.
We know that:
$9 = 3^2 \quad \text{and} \quad 243 = 3^5.$
Substitute these into the given expression:
$\frac{9^{3y}}{243} = \frac{(3^2)^{3y}}{3^5}.$
Step 2: Simplify the numerator.
Using the power rule $(a^m)^n = a^{mn}$, we get:
$(3^2)^{3y} = 3^{6y}.$
Thus:
$\frac{9^{3y}}{243} = \frac{3^{6y}}{3^5}.$
Step 3: Simplify the division.
Using the rule $\frac{a^m}{a^n} = a^{m-n}$, we have:
$\frac{3^{6y}}{3^5} = 3^{6y-5}.$
Hence:
$\frac{9^{3y}}{243} = 3^{6y-5}.$
(b) Solve the simultaneous equations
The given equations are:
$\frac{9^{3y}}{243} = 27^{x-2},$
$\log_{10} \sqrt{6xy} = \log_4 2.$
Step 1: Rewrite the first equation.
From part (a), we know:
$\frac{9^{3y}}{243} = 3^{6y-5}.$
Rewrite $27$ as $3^3$:
$27^{x-2} = (3^3)^{x-2}.$
Simplify using the power rule:
$27^{x-2} = 3^{3(x-2)} = 3^{3x-6}.$
Equating the exponents:
$6y - 5 = 3x - 6.$
Rearrange:
$6y = 3x - 1.$
Simplify:
$2y = x - \frac{1}{3}.$
$x = 2y + \frac{1}{3}.$
Step 2: Rewrite the second equation.
The second equation is:
$\log_{10} \sqrt{6xy} = \log_4 2.$
Simplify $\sqrt{6xy}$ as $(6xy)^{1/2}$:
$\log_{10} \sqrt{6xy} = \frac{1}{2} \log_{10}(6xy).$
Thus:
$\frac{1}{2} \log_{10}(6xy) = \log_4 2.$
Multiply through by 2:
$\log_{10}(6xy) = 2 \log_4 2.$
Using the change of base formula $\log_a b = \frac{\log_c b}{\log_c a}$, rewrite $\log_4 2$:
$\log_4 2 = \frac{\log_{10} 2}{\log_{10} 4}.$
Since $\log_{10} 4 = \log_{10}(2^2) = 2\log_{10} 2$, we have:
$\log_4 2 = \frac{\log_{10} 2}{2 \log_{10} 2} = \frac{1}{2}.$
Substitute this back:
$\log_{10}(6xy) = 2 \cdot \frac{1}{2} = 1.$
Thus:
$\log_{10}(6xy) = 1.$
Step 3: Solve for $6xy$.
The equation $\log_{10}(6xy) = 1$ implies:
$6xy = 10^1 = 10.$
Simplify:
$xy = \frac{10}{6} = \frac{5}{3}.$
Step 4: Substitute $x = 2y + \frac{1}{3}$ into $xy = \frac{5}{3}$.
Substitute $x$ into $xy = \frac{5}{3}$:
$\left(2y + \frac{1}{3}\right)y = \frac{5}{3}.$
Expand:
$2y^2 + \frac{1}{3}y = \frac{5}{3}.$
Multiply through by 3:
$6y^2 + y = 5.$
Rearrange:
$6y^2 + y - 5 = 0.$
Step 5: Solve the quadratic equation.
Using the quadratic formula:
$y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a},$
where $a = 6$, $b = 1$, and $c = -5$:
$y = \frac{-1 \pm \sqrt{1^2 - 4(6)(-5)}}{2(6)}.$
Simplify:
$y = \frac{-1 \pm \sqrt{1 + 120}}{12}.$
$y = \frac{-1 \pm \sqrt{121}}{12}.$
$y = \frac{-1 \pm 11}{12}.$
Solve for $y$:
$y = \frac{10}{12} = \frac{5}{6}, \quad \text{or} \quad y = \frac{-12}{12} = -1.$
$y = \frac{5}{6} \text{ (or) } y=-1.$
Step 6: Solve for $x$, when $y=\frac{5}{6}$.
Substitute $y = \frac{5}{6}$ into $x = 2y + \frac{1}{3}$:
$x = 2\left(\frac{5}{6}\right) + \frac{1}{3}.$
Simplify:
$x = \frac{10}{6} + \frac{2}{6} = \frac{12}{6} = 2.$
Step 7: Solve for $x$, when $y=-1$.
Substitute $y = -1$ into the equation $x = 2y + \frac{1}{3}$:
$x = 2(-1) + \frac{1}{3}.$
Simplify:
$x = -2 + \frac{1}{3}.$
Combine the terms:
$x = \frac{-6}{3} + \frac{1}{3} = \frac{-5}{3}.$
Final Answer:
$x = 2, \quad y = \frac{5}{6}.$
$x = \frac{-5}{3}, \quad y = -1.$
Post a Comment