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# system

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Let a, b, c be positive real numbers such that bc = a/2, ca = 8b, and ab = 128c.  Find a + b + c.

May 13, 2020

#1
+498
+1

If you multiply all three of our given expressions together, you get:

\((abc)^2 = (1/2 * 8 * 128)abc\)

Dividing by abc on both sides, we then get:
\(abc = 64 * 8 = 512\)

If we look back to our first expression \(bc = a/2\), we can multiply a on both sides to get a known value(abc).

This gives us:

\(abc = a^2/2\)

\(512 = a^2/2\)

\(a^2=1024\)

Realize that 1024 = 210, so the square root of it is simply 25 = 32

\(a = 2^5 = 32\)

Repeat the same steps for our second equation by multiplying by b on both sides, and we get:
\(abc = 8b^2\)

\(512 = 8b^2\)

\(64 = b^2\)

\(b = 8\)

\(abc = 512\)

Since a = 32 and b = 8

That only leaves us with c, which is then:

\(512/(32*8) = 2\)

a = 32

b = 8

c = 2

a+b+c = 32 + 8 + 2 = 42

May 13, 2020

#1
+498
+1

If you multiply all three of our given expressions together, you get:

\((abc)^2 = (1/2 * 8 * 128)abc\)

Dividing by abc on both sides, we then get:
\(abc = 64 * 8 = 512\)

If we look back to our first expression \(bc = a/2\), we can multiply a on both sides to get a known value(abc).

This gives us:

\(abc = a^2/2\)

\(512 = a^2/2\)

\(a^2=1024\)

Realize that 1024 = 210, so the square root of it is simply 25 = 32

\(a = 2^5 = 32\)

Repeat the same steps for our second equation by multiplying by b on both sides, and we get:
\(abc = 8b^2\)

\(512 = 8b^2\)

\(64 = b^2\)

\(b = 8\)

\(abc = 512\)

Since a = 32 and b = 8

That only leaves us with c, which is then:

\(512/(32*8) = 2\)

a = 32

b = 8

c = 2

a+b+c = 32 + 8 + 2 = 42

jfan17 May 13, 2020
#2
+114141
+1

Very impressive, jfan  !!!!!

CPhill  May 13, 2020