Solving for d_2: Simple Algebra Help

  • Thread starter Thread starter DB
  • Start date Start date
  • Tags Tags
    Algebra
AI Thread Summary
To solve for d_2 in the equation 336/81 = (175/100) * (1/(d_2/0.1)^3), the key is to manipulate the equation using algebraic principles. By recognizing that if a = 1/b^3, then b^3 = 1/a, the next steps involve isolating d_2. The correct approach leads to d_2 being expressed as [4.1 * 10^{-4}]^(1/3). Keeping constants in fraction form until the final steps simplifies the cube root calculation. This method ultimately yields a more accurate result for d_2.
DB
Messages
501
Reaction score
0
to solve for d_2:
\frac{336}{81}=\frac{175}{100}*\frac{1}{[\frac{d_2}{0.1}]^3}
\sim 2.4=\frac{1}{[\frac{d_2}{0.1}]^3}

then i don't know wat to do :mad: i know its simple algebra, i just don't see it,.
 
Physics news on Phys.org
If a = \frac{1}{b^3}
Then b^3 = \frac{1}{a}

Use this idea and keep going.
 
thnx 4 all the help doc,
i get
d_2=0.041^{\frac{1}{3}}
 
DB said:
i get
d_2=0.041^{\frac{1}{3}}
That doesn't seem right. Consider the following:
[\frac{d_2}{0.1}]^3 = \frac{d_2^3}{0.001}
 
o i see wat i did, so d_2=[4.1*10^{-4}]^\frac{1}{3}?
 
Much better.
 
Even better if you don't convert your constants to a decimal so soon.
Keep them as fractions and you should find you can easily take their cube root later. :wink:
 
Back
Top