Question

Log_(0, 5)(4x - 3) >= log_(0, 5)(x + 1)

Answer 1

Answer: Solution

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−log

2

​

(4−x)≥−log

2

​

2+log

2

​

(x−1)

log

2

​

2−log

2

​

(4−x)≥log

2

​

(x−1)

Taking antilog, we get,

4−x

2

​

≥x−1

4−x

2−(4−x)(x−1)

​

≥0

4x

2−(4x−4−x

2

+x)

​

≥0

x−4

2−5x+x

2

+4

​

≥0

x−4

x

2

−5x+6

​

≤0

(x−4)

(x−3)(x−2)

​

≤0

therefore,

xε(−∞,2]∪[3,4)

Now,asx<4andx>1

xε(1,4)

so,wecansaythat,

xε(1,2]∪[3,4)

Explanation: