urgent: log questions (1 Viewer)

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I got an exam tommorrow

1. Solve 2^x.2^(x+1) = 14 for x

and

2. Simplify ln3x + ln4 - ln2x

Thanks

 

[insert-username]

1. Solve 2^x.2^(x+1) = 14 for x

2^{x^2 + x + 1} = 14 (index law)

x² + x + 1 = log₂14

x² + x + 1 - log₂14 = 0

x² + x + log₂2 - log₂14 = 0

x² + x + log₂1/7 = 0

Use quadratic formula from there to solve for x.

2. Simplify ln3x + ln4 - ln2x

Remember your log laws: when you subtract logs of the same base, you divide, and when you add logs of the same base, you multiply.

ln3x + ln4 - ln2x = ln 12x - ln 2x

= ln 6


I_F

 

[Riviet]

For the first one, I think he meant:
2^x.2^x+1=14
2^2x+1=14
log₂2^2x+1=log₂14
(2x+1).log₂2=log₂14
x=(log₂14 - 1)/2
Use a calculator if necessary.

 

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much <3 to insert-username and riviet