Detailed Solution for Exponential Equation
Understanding the Problem
We are given an exponential equation involving fractions: $ \left(\frac{7}{11}\right)^{k-5} = \left(\frac{11}{7}\right)^{k-9} $ Our goal is to find the value of $k$ and then calculate $2^k$.
Applying Exponent Rules
To solve this equation, we need to make the bases on both sides the same. We observe that the base on the right side, $\frac{11}{7}$, is the reciprocal of the base on the left side, $\frac{7}{11}$.
We know that the reciprocal of a number can be represented using a negative exponent. Specifically, $a^{-n} = \frac{1}{a^n}$. Therefore, we can rewrite $\frac{11}{7}$ as:
$ \frac{11}{7} = \left(\frac{7}{11}\right)^{-1} $ Now, substitute this back into the original equation: $ \left(\frac{7}{11}\right)^{k-5} = \left(\left(\frac{7}{11}\right)^{-1}\right)^{k-9} $ Using the power of a power rule, $(a^m)^n = a^{m \times n}$, we simplify the right side: $ \left(\frac{7}{11}\right)^{k-5} = \left(\frac{7}{11}\right)^{-(k-9)} $Solving for k
Since the bases are now the same ($\frac{7}{11}$), the exponents must be equal:
$ k-5 = -(k-9) $ Now, we solve this linear equation for $k$:- Distribute the negative sign on the right side: $ k-5 = -k + 9 $
- Add $k$ to both sides of the equation: $ k + k - 5 = -k + k + 9 $ $ 2k - 5 = 9 $
- Add 5 to both sides of the equation: $ 2k - 5 + 5 = 9 + 5 $ $ 2k = 14 $
- Divide by 2 to find the value of $k$: $ k = \frac{14}{2} $ $ k = 7 $
Calculating the Final Value
The question asks for the value of $2^k$. We found that $k=7$. Substitute this value into the expression $2^k$:
$ 2^k = 2^7 $Now, calculate $2^7$:
$ 2^7 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 128 $Final Answer
The value of $2^k$ is 128.
Important Questions from Algebra (Notes)
- If $(y-12) = 4\sqrt{5}$, then find the value of $\sqrt{y-3} - \frac{1}{\sqrt{y-3}}$.
- If $x^2 + \frac{1}{x^2} = 16$ and $x \neq 0$, then what is the value of $x^4 + \frac{1}{x^4}$?
- In the expansion of (x + 9)(x - 6)(x + 5), what is the coefficient of x?
- Find the value of $\frac{x+3}{x^2-2x} \times \frac{2x-1}{x^2+2x+4} \times \frac{x^4-8x}{2x^2+5x-3}$
- The roots of the equation $ax^3-24x^2+188x-480=0$ are three consecutive even natural numbers. The value of a is _____.
