Question

Suppose that S varies directly as the 3​/5 power of​ T, and that Sequals24 when Tequals32. Find S when Tequals243.
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Part 1
Sequals
  
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Answer 1

Since \( S \) varies directly as the \( \frac{3}{5} \) power of \( T \), we can express this relationship as:

\[
S = k \cdot T^{\frac{3}{5}}
\]

where \( k \) is a constant.

We know that \( S = 24 \) when \( T = 32 \). We can use this information to find the constant \( k \).

Substituting the values into the equation:

\[
24 = k \cdot (32)^{\frac{3}{5}}
\]

Next, we need to calculate \( (32)^{\frac{3}{5}} \):

1. Calculate \( 32^{\frac{1}{5}} \):
\[
32^{\frac{1}{5}} = 2
\]
(since \( 2^5 = 32 \))

2. Now raise it to the 3rd power:
\[
(32^{\frac{1}{5}})^{3} = 2^3 = 8
\]

Now substitute back into the equation:

\[
24 = k \cdot 8
\]

To find \( k \):

\[
k = \frac{24}{8} = 3
\]

Now we have the relationship:

\[
S = 3 \cdot T^{\frac{3}{5}}
\]

Next, we want to find \( S \) when \( T = 243 \):

\[
S = 3 \cdot (243)^{\frac{3}{5}}
\]

To calculate \( (243)^{\frac{3}{5}} \):

1. Calculate \( 243^{\frac{1}{5}} \):
\[
243^{\frac{1}{5}} = 3
\]
(since \( 3^5 = 243 \))

2. Raise it to the 3rd power:
\[
(243^{\frac{1}{5}})^{3} = 3^3 = 27
\]

Now substitute back into the equation for \( S \):

\[
S = 3 \cdot 27 = 81
\]

So, when \( T = 243 \), \( S \) equals \( \boxed{81} \).