Janek191:
a) [ x + 3 > 0 i x − 1 > 0 ] ⇒ x > 1
log
4 ( x + 3) − 2 = log
4 ( x − 1) − log
4 8
log
4 ( x + 3) − log
4 ( x − 1) = 2 − log
22 2
3
| | x + 3 | | 1 | |
log4 |
| = 2 − 3* |
| log2 2 |
| | x − 1 | | 2 | |
| | x + 3 | | 1 | |
log4 |
| = 2 − 1,5 = |
| |
| | x −1 | | 2 | |
x + 3 = 2*( x − 1)
x + 3 = 2 x − 2
3 + 2 = 2 x − x
5 = x
x = 5
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Janek191:
c) x + 10 > 0 i 2 x − 1 > 0 i 21 x − 20 > 0
| | 20 | |
x > − 10 i x > 0,5 i x > |
| |
| | 21 | |
−−−−−−−−−−−−−−−−
log 5 + log ( x + 10 ) = 1 − log ( 2 x − 1) + log ( 21 x − 20 )
log 5*( x + 10) − log 10 = log ( 21 x − 20) − log ( 2 x − 1)
| | 5(x + 10) | | 21 x − 20 | |
log |
| = log |
| |
| | 10 | | 2 x − 1 | |
| x + 10 | | 21 x − 20 | |
| = |
| |
| 2 | | 2 x − 1 | |
2*( 21 x − 20 ) = ( x + 10)*( 2 x − 1)
42 x − 40 = 2 x
2 + 19 x − 10
2 x
2 − 23 x + 30 = 0
Δ = 529 − 4*2*30 = 529 − 240 = 289
√Δ = 17
| | 23 − 17 | | 23 + 17 | |
x = |
| = 1,5 lub x = |
| = 10 |
| | 4 | | 4 | |
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