Think fast

Algebra Level 2

Find the number of integral values of x satisfying the inequality l o g 4 2 x − 1 x + 1 < = c o s 4 Π 3 log_{4}\frac{2x-1}{x+1}<=cos\frac{4Π}{3}


The answer is 1.

Harsh Poddar by harsh poddar , 22, India

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

3 solutions

log ⁡ 4 2 x − 1 x + 1 ≤ cos ⁡ 4 π 3 = − 1 2 ⇒ 2 x − 1 x + 1 ≤ 4 − 1 2 = 1 2 4 x − 2 ≤ x + 1 3 x ≤ 3 ⇒ x ≤ 1 \begin{aligned} \log_4 \frac{2x-1}{x+1} & \le \cos{\frac{4\pi}{3}} = -\frac{1}{2} \\ \Rightarrow \frac{2x-1}{x+1} & \le 4^{-\frac{1}{2}} = \frac{1}{2} \\ 4x - 2 & \le x + 1 \\ 3x & \le 3 \\ \Rightarrow x & \le 1 \end{aligned}

Since log ⁡ 4 2 x − 1 x + 1 ∈ ( − 2 , ∞ ) \log_4 \dfrac{2x-1}{x+1} \in (-2, \infty) , there is only 1 \boxed{1} integer value of x = 1 x=1 satisfying the inequality.

5 Helpful 0 Interesting 0 Brilliant 0 Confused
Akshay Joshi
Aug 8, 2015

4^-0.5=(2x-1)/(x+1);; 1/2=(2x-1)/(x+1);; x=1---ans

0 Helpful 0 Interesting 0 Brilliant 0 Confused
Rajkumar Seth
Aug 8, 2015

cos(4pi/3)=-1/2 2x-1/x+1 <=4^(-1/2)=1/2 4x-2<=x+1 x<=1

0 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...