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What is the true solution to 3 In 2+In 8 = 2 In(4x)?
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TNT 00

User GManNickG
by
6.6k points

1 Answer

3 votes

Answer:

x = 2

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

User Ekerner
by
6.5k points
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