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3 years ago

Given that 2^A×3^B×5^13=20^D×18^12, where A,B, and D are postive integers, compute A+B+D.

Mathematics

1 answer:

Taya2010 [7]3 years ago

8 0

Answer: 75

Step-by-step explanation:

$2^A3^B5^{13}=20^D18^{12}$

⇒ $2^A3^B5^{13}=(2^2\cdot5^1)^D(2^1\cdot3^2)^{12}$

⇒ $2^A3^B5^{13}=(2^{2D}\cdot5^D)(2^{12}\cdot3^{24})$

⇒ $2^A3^B5^{13}=2^{2D+12}\cdot3^{24}\cdot5^D$

Now compare the like bases:

$2^A=2^{2D+12}$ ⇒ A = 2D + 12

$3^B=3^{24}$ ⇒ B = 24

$5^{13}=5^D$ ⇒ D = 13

Next, let's solve for A:

A = 2D + 12

= 2(13) + 12

= 26 + 12

= 38

LAST STEP: Find the sum of A, B, and D

S = A + B + D

= 38 + 24 + 13

= 75

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