Solve for k: 8√8 = 8^k - Easy Method Explained

  • Thread starter Thread starter thomas49th
  • Start date Start date
AI Thread Summary
To solve for k in the equation 8√8 = 8^k, it is necessary to express both sides in terms of the same base. The left side can be rewritten as 8^(3/2) since √8 equals 8^(1/2). Therefore, k is determined to be 3/2. The method involves converting the square root and simplifying the exponents accordingly. This approach clarifies the solution without the need for trial and error.
thomas49th
Messages
645
Reaction score
0
8\sqrt{8} can be written in the form 8^{k}

Find the value of k

I'm positive is a really easy question I though that k = 1/8, but that is wrong.
What is the method to answer this question.. I used trial and error and got k = 3/2, but what is the proper way?

Thanks
 
Physics news on Phys.org
Well what is a if \sqrt{8} = 8^{a}?

What is b if 8 = 8^{b}?

What is 8^a\cdot8^b?
 
8^{\frac{2}{2}} x 8^{\frac{1}{2}}

= 8^{\frac{3}{2}}

cheerz
 
thomas49th said:
8^{\frac{2}{2}} x 8^{\frac{1}{2}}

= 8^{\frac{3}{2}}

cheerz
Pleasure :smile:
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
Back
Top