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50 POINTS PLEASE HELP

50 POINTS PLEASE HELP-example-1

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3 votes

Answer:

2a+3b is 13

Explanation:

User Floele
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We know that a^2b^3=108. We can factor 108 into its prime factors: 108 = 2^2 x 3^3. We can then rewrite a^2b^3 as (a^2)(b^3) = (2^2)(3^3).

Since a and b are positive integers, we can see that a^2 must be one of the factors of 2^2, and b^3 must be one of the factors of 3^3. The only pairs of factors that multiply to 108 and satisfy these conditions are (a^2, b^3) = (2^2, 3^3) and (a^2, b^3) = (1, 108).

The first pair gives us a=2 and b=3, while the second pair gives us a=1 and b=108. Since a and b are positive integers, we can discard the second solution, so we have a=2 and b=3.

Finally, we can substitute these values into the expression 2a + 3b to get:

2a + 3b = 2(2) + 3(3) = 4 + 9 = 13.

Therefore, the value of 2a + 3b is 13.
User Tom Feiner
by
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