The sum of all real values of k for which (81)k×(327681)31=(81)×(327681)k1 is
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Official Correct Answer: B. The equation given is:
$$
\left( \dfrac{1}{8} \right)^k \times \left( \dfrac{1}{32768} \right)^{\frac{1}{3}} = \left( \dfrac{1}{8} \right) \times \left( \dfrac{1}{32768} \right)^{\frac{1}{k}}
$$
We find all real values of \( k \) that satisfy this equation and then sum them up. The options are given, and the correct answer is supposed to be B, which is \( -\frac{2}{3} \). But I need to figure out how to get there step by step.
First, Expressing all the numbers as powers of smaller bases to make it easier to handle the exponents. Note that 8 and 32768 are both powers of 2. Let me confirm that.
8 is \( 2^3 \), right? Because \( 2 \times 2 \times 2 = 8 \). So, \( \frac{1}{8} \) is \( 2^{-3} \).
Now, 32768. Hmm, I think that's a larger power of 2. Calculate:
\( 2^{10} = 1024 \)
\( 2^{15} = 32768 \). Yes, because \( 2^{10} \times 2^5 = 1024 \times 32 = 32768 \). So, 32768 is \( 2^{15} \), which means \( \frac{1}{32768} = 2^{-15} \).
So, substituting these into the equation, I can rewrite everything in terms of base 2. Let's do that.
Left side:
$$
\left( 2^{-3} \right)^k \times \left( 2^{-15} \right)^{\frac{1}{3}}
$$
Right side:
$$
\left( 2^{-3} \right) \times \left( 2^{-15} \right)^{\frac{1}{k}}
$$
Now, using the exponent rule \( (a^m)^n = a^{mn} \), I can simplify both sides.
Starting with the left side:
First term: \( (2^{-3})^k = 2^{-3k} \)
Second term: \( (2^{-15})^{1/3} = 2^{-15 \times \frac{1}{3}} = 2^{-5} \)
So, the left side becomes:
$$
2^{-3k} \times 2^{-5} = 2^{-3k - 5}
$$
Because when you multiply exponents with the same base, you add the exponents: \( a^m \times a^n = a^{m+n} \).
Now, the right side:
First term: \( 2^{-3} \)
Second term: \( (2^{-15})^{1/k} = 2^{-15 \times \frac{1}{k}} = 2^{-15/k} \)
So, the right side becomes:
$$
2^{-3} \times 2^{-15/k} = 2^{-3 - 15/k}
$$
Again, using the same exponent rule as before.
Now, since both sides have the same base, 2, their exponents must be equal for the equation to hold. So, I can set the exponents equal to each other:
$$
-3k - 5 = -3 - \frac{15}{k}
$$
Now, I need to solve this equation for \( k \). Let's write that out:
$$
-3k - 5 = -3 - \frac{15}{k}
$$
Hmm, this looks a bit complicated because there's a \( k \) in the denominator. Maybe I can multiply both sides by \( k \) to eliminate the fraction. Let's try that.
Multiplying every term by \( k \):
Left side: \( -3k \times k - 5 \times k = -3k^2 -5k \)
Right side: \( -3 \times k - \frac{15}{k} \times k = -3k -15 \)
So, the equation becomes:
$$
-3k^2 -5k = -3k -15
$$
Now, let's bring all terms to one side to set the equation to zero. I'll add \( 3k + 15 \) to both sides:
Left side: \( -3k^2 -5k + 3k +15 \)
Right side: \( -3k -15 + 3k +15 = 0 \)
Simplify the left side:
Combine like terms: \( -5k + 3k = -2k \)
So, left side becomes:
$$
-3k^2 -2k +15 = 0
$$
Hmm, that's a quadratic equation. Let me write it as:
$$
-3k^2 -2k +15 = 0
$$
It's better to have the coefficient of \( k^2 \) positive, so I'll multiply both sides by -1:
$$
3k^2 + 2k -15 = 0
$$
Now, I have a quadratic equation in standard form \( ax^2 + bx + c = 0 \), where \( a = 3 \), \( b = 2 \), and \( c = -15 \).
I can solve this using the quadratic formula:
$$
k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
$$
Plugging in the values:
$$
k = \frac{-2 \pm \sqrt{(2)^2 - 4 \times 3 \times (-15)}}{2 \times 3}
$$
Calculate the discriminant first:
\( b^2 - 4ac = 4 - 4 \times 3 \times (-15) = 4 + 180 = 184 \)
So, the square root part is \( \sqrt{184} \). if I can simplify that.
184 divided by 4 is 46, so \( \sqrt{184} = \sqrt{4 \times 46} = 2\sqrt{46} \).
So, plugging back into the formula:
$$
k = \frac{-2 \pm 2\sqrt{46}}{6}
$$
Factoring out a