Question

Find A and B where 3^12 x 6^8 = 3^4 x 4^B​

Answer 1

check detail for the solution

Answer 2

The value of A is 20 and B is 4.

Given:

$3^{12} \times 6^8 = 3^A \times 4^B$

To Find:

The values of A and B

Solution:

To find the value of A and B, we have to make the base, on both sides of the equation, the same.

$\implies 3^{12} \times 6^8 = 3^A \times 4^B\\\\\implies 3^{12} \times (3 \times 2)^8 = 3^A \times 4^B\\\\\implies 3^{12} \times 3^8 \times 2^8 = 3^A \times 4^B\\\\\implies 3^{20} \times 2^8 = 3^A \times 4^B\\\\\implies 3^{20} \times (2 \times2 \times2 \times2 \times2 \times2 \times2 \times2) = 3^A \times 4^B\\\\\implies 3^{20} \times (4 \times4 \times4 \times4) = 3^A \times 4^B\\\\\implies 3^{20} \times 4^4 = 3^A \times 4^B$

Now, comparing the LHS and RHS we get,

⇒ A = 20, B = 4        (Ans.)

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