1. ## trignometry

solve the eqn:

1/log(base 4) (x+1)/(x+2) < 1/log(base 4) (x+3)

regret me
the topic is LOGARITHMS

2. Hello, grgrsanjay!

I have a very long solution . . . and it may be incomplete.
Someone check it carefully . . . please!

$\text{Solve: }\;\dfrac{1}{\log_4\!\left(\frac{x+1}{x+2}\right)} \;<\; \dfrac{1}{\log_4(x+3)}\;\;[1]$

We begin with: . $1 \:<\:2$

Add $x\!:\;\;x+1 \:<\:x+2\;\;[2]$

Domain: $x \,>\,-1 \quad\Rightarrow\quad x+1 \,>\,0 \quad\Rightarrow\quad x+2 \,>\,0$

Divide [2] by positive $x+2\!:$ . $\dfrac{x+1}{x+2} \:<\:1$

Take logs, base 4: . $\log_4\!\left(\dfrac{x+1}{x+2}\right)\:<\:\log_4(1 ) \:=\:0$

That is: . $\log_4\!\left(\dfrac{x+1}{x+2}\right)$ .is negative.

Multiply [1] by negative $\log_4\!\left(\frac{x+1}{x+2}\right)\!:$ . $1 \;>\;\dfrac{\log_4\!\left(\frac{x+1}{x+2}\right)}{ \log_4(x+3)}\;\;[3]$

Since $x \,>\, -1$, then $x+3 \,>\,2 \quad\Rightarrow\quad \log_4(x+3) \:>\:0$

Multiply [3] by positive $\log_4(x+3)\!:$ . $\log_4(x+3) \;>\;\log_4\!\left(\dfrac{x+1}{x+2}\right)$

Take anti-logs: . $x+3\;>\;\dfrac{x+1}{x+2} \quad\Rightarrow\quad x^2 + 4x + 5 \;>\;0$

This is an up-opening parabola with its vertex at (-2, 1).
. . Hence, it is always above the $x$-axis.

The inequality is true for all $x$ in the domain: . $x \:>\:-1$