Question

Logarytmia

1)

2)

3)

4)

5)

6)

7)

8)

Answer 1

1)  log 3 [ x + 2 ] = 4  <=> x+2 = 3^4 =81  <=> x = 79

x = 79

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2)  log (3^x) [ 9 ] = - 1  <=> ( 3 ^x)^(-1) = 9  <=> 3^ (-x) = 3^2 <=> -x = 2 <=> x = - 2

x = - 2

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3)  log 4 [ x - 5 ] = 4  <=> x - 5 = 4^4 = 256 <=> x = 261

x = 261

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4)   log 3 [ x - 2 ] + log 3 [ x ] = 0

założenie:  x > 2   i  x > 0 =>  x > 2

log 3 [ x - 2] = - log 3 [ x ]

log 3 [x - 2] = log 3 [ x ^(-1)]

log 3 [ x - 2 ] = log 3 [ 1/x ]

x - 2 = 1/x    / * x

x^2 - 2x - 1 = 0

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delta = (-2)^2 - 4*1*(-1) = 4 + 4 = 4*2

p (delty ) = 2 p(2)

x = [ 2 - 2 p(2)]/2 = 1 - p(2) < 0  - odpada

x = [ 2 + 2 p(2)]/2 = 1 + p(2) > 2

x = 1 + p(2)

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5)

ln [x + 1] +  ln [x + 2 ] = 0

założenie: x> - 1  i  x > - 2  czyli  x > - 1

ln[ x + 1 ] = - ln [ x + 2]

ln [ x + 1 ] = ln[ 1/(x + 2 )]

x + 1 = 1 /(x + 2)   / * x+ 2

(x +1)*(x + 2) = 1

x^2 +2x + x + 2 - 1 = 0

x^2 + 3 x + 1 = 0

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delta = 3^2 - 4*1*1 = 9 - 4 = 5

p (delty) = p(5)

x = [ - 3 - p(5)]/2  <  - 1 - odpada

x = [ - 3 + p(5)]/2  > - 1

Odp. x = - 1,5 + 0,5 p(5)

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6)

log 4 [ x ] + 4 log 4 [ x - 5 ]

coś brakuje

7)

log 1/5 [ x + 4 ] + log 1/5 [ x ] > = log 1/5 [ 3 ] ; założenie: x > 0

log 1/5 [ x^2 + 4 x ] > = log 1/5 [ 3 ]

 x^2 + 4 x < = 3

x^2 + 4 x -3 < = 0

delta = 4^2 - 4*1*(-3) = 16 + 12 = 28 = 4*7

p (delty) = 2 p(7)

x = [ - 4 - 2 p(7)]/ 2 = - 2 -  p(7) < 0  - odpada

x = [ - 4 + 2 p(7)]/2 = - 2 + p(7) > 0

Odp. x = -2 + p(7)

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8)

4 ln 2 - ln x + ln 3

coś brakuje