Q) The solution of Logarithmic inequality log2x-1+logx-9 1 (A) (0, 13) (B) (13, ∞) (C) (- 13, 13) (D) - Maths - Relations and Functions

Question

Q) The solution of Logarithmic inequality log 2 x - 1 + log x - 9
> 1
(A) (0, 13)
(B) (13, ∞ )
(C) (- 13, 13)
(D) None of these

Aarushi Mishra · Accepted Answer

Assuming hte base of the lograithm to be 10log10 2x-1+log10 x-9>1qunatity inside square root should be non-negative
 Also quantity inside log should be non-zero2x-1>0 and x-9>0x>12 and x>9So x>9Use log m+log n=log mnlog10 2x-1+log10 x-9>1log10 2x-1x-9>1Using logarithmic inequality for base>1
 logab>c⇒b>ac 2x-1x-9>102x-1x-9>1002x2-19x+9>1002x2-19x-91>0Note: root so of the equation 2x2-19x-91=0, are x=19±192+8×914x=19±361+8×914=19±10894=cx=19+334 or x=19-334x=524 or x=-144x=13 or x=-72so x-13 and x+72 are factors of 2x2-19x-91x-13 x+72>0x-13 x--72>0Solution of the equation x-ax-b>0, where a<b, is given by x∈-∞,a∪b, ∞x∈-∞,-72∪13, ∞But x>9therefore,x∈13, ∞

Q). The solution of Logarithmic inequality log 2 x - 1 + log x - 9 > 1 (A) (0, 13) (B) (13, ∞ ) (C) (- 13, 13) (D) None of these

Q). The solution of Logarithmic inequality $\log \sqrt{(2 x - 1)} + \log \sqrt{(x - 9)} > 1$
(A) (0, 13)
(B) (13, $\infty$ )
(C) (- 13, 13)
(D) None of these