Question

What is the true solution to 3 In 2+In 8 = 2 In(4x)? 60f Š O OOO TNT 00

Answer 1

Answer:

  x = 2

Step-by-step explanation:

Taking antilogs, you have ...

  2³ × 8 = (4x)²

  64 = 16x²

  x = √(64/16) = √4

  x = 2 . . . . . . . . (the negative square root is not a solution)

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You can also work more directly with the logs, if you like.

  3·ln(2) +ln(2³) = 2ln(2²x) . . . . . . . . . . . write 4 and 8 as powers of 2

  3·ln(2) +3·ln(2) = 2(2·ln(2) +ln(x)) . . . . use rules of logs to move exponents

  6·ln(2) = 4·ln(2) +2·ln(x) . . . . . . . . . . . . simplify

  2·ln(2) = 2·ln(x) . . . . . . . . . . . subtract 4ln(2)

  ln(2) = ln(x) . . . . . . . . . . . . . . divide by 2

  2 = x . . . . . . . . . . . . . . . . . . . take the antilogs