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Property 3

If log02x -...

Question

If ${log}_{02} (x - 1) > {log}_{004} (x + 5)$ then

A

- 1 < x < 4

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B

2 < x < 3

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C

1 < x < 4

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D

1 < x < 3

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Solution

The correct option is D $1 < x < 4$
${log}_{0.2} (x - 1) > {log}_{{0.2}^{2}} (x + 5) \Rightarrow {log}_{0.2} (x - 1) > \frac{1}{2} {log}_{0.2} (x + 5)$
$\Rightarrow (x - 1)^{2} < (x + 5) \Rightarrow x^{2} - 3 x - 4 < 0 \Rightarrow - 1 < x < 4$ (but $x > 1$ )
$\Rightarrow 1 < x < 4$

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Similar questions

{log}_{0.2} (x - 1) > {log}_{0.04} (x + 5)

If f(x+1/x)=x²+1/x², then the value of f(3/2) is:

a) 5/4

b) 4/5

c) 1/4

d) -1/4

\frac{3 x^{2} - 2 x + 4}{(x - 1)^{2}} = \frac{A_{1}}{x + 1} + \frac{A_{2}}{(x + 1)^{2}} + \frac{A_{3}}{(x + 1)^{3}} + \frac{A_{4}}{(x + 1)^{4}} + \frac{A_{5}}{(x + 1)^{5}} + \frac{A_{6}}{(x + 1)^{6}}

(A_{1} + A_{3} + A_{5}, A_{2} + A_{4} + A_{6}) =

x \neq - 1

, then find the equation of

\frac{x^{5} + x^{4} + x^{3} + x^{2}}{x^{3} + x^{2} + x + 1} = 1

x^{3} - \frac{1}{x^{3}} = 14

x - \frac{1}{x} =

(a) 5

(b) 4

(c) 3

(d) 2

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