# trignometry

• Aug 5th 2010, 07:07 AM
grgrsanjay
trignometry
solve the eqn:

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

regret me
the topic is LOGARITHMS
• Aug 5th 2010, 08:42 AM
Soroban
Hello, grgrsanjay!

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

Quote:

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

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

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

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

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

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

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

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

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

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

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

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

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