Solving log₂√(2x-1) < log₄x inequality

[Helly123]

Homework Statement

$$\log_2 \sqrt{2x-1} < \log_4 x$$

Homework Equations

The Attempt at a Solution

$$\log_2 \sqrt{2x-1} < \log_4 x$$
$$\log_2 {2x-1}^{1/2} < \log_2 \sqrt{x}$$
$$1/2 \log_2 {(2x-1)} - \log_2 \sqrt{x} < 0$$
$$ 1/2 \log_2 {\frac{(2x-1)}{ x}} < 0$$
$$ -1/2 \log_2 {(2x-1) x} < 0$$
$$ 1/2 \log_2 {(2x^2-x)} > 0$$
$$ \log_2 {\sqrt{(2x^2-x)}} >\log_2 {1} $$
$$2x^2 - x > 1$$
$$2x^2 - x - 1 = 0$$
$$(2x + 1)(x -1) = 0$$
$$x < -1/2 \ \ or \ \ x > 1$$

but I get wrong answers. How to find the right answer? can you help me?

 

[ehild]

Homework Statement

$$\log_2 \sqrt{2x-1} < \log_4 x$$

Homework Equations

The Attempt at a Solution

$$\log_2 {2x-1}^{1/2} < \log_2 \sqrt{x}$$

You miss parentheses on the left side and the base of logatrithm is 4 instead of 2 on the right side.

 

[Helly123]

You miss parentheses on the left side and the base of logatrithm is 4 instead of 2 on the right side.

That's exactly what the question is..

 

[ehild]

That's exactly what the question is..

No. The question is :
$\log_2 \sqrt{2x-1} < \log_4 x$

 

[Marc Rindermann]

I can't really follow all your steps because it's almost 2am over here and it's been a long day. However, your second line can be written as

$\frac12 \log_2(2x - 1) < \frac12 \log_2(x)$

from here it should be solved in 3 steps

 

[Helly123]

No. The question is :
$\log_2 \sqrt{2x-1} < \log_4 x$

ive changed it

 

[Helly123]

I can't really follow all your steps because it's almost 2am over here and it's been a long day. However, your second line can be written as

$\frac12 \log_2(2x - 1) < \frac12 \log_2(x)$

from here it should be solved in 3 steps

I see.. that's true. Thank you