# logarithms

#### grgrsanjay

solve the eqn:

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

#### Soroban

MHF Hall of Honor
Hello, grgrsanjay!

I already solved this somewhere . . . Here is it again.

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

Note that the domain is: .$$\displaystyle x > -1$$

For $$\displaystyle x > -1,$$ .$$\displaystyle \dfrac{x+1}{x+2} \:<\:1$$

. . Hence: .$$\displaystyle \log_4\!\left(\dfrac{x+1}{x+2}\right) \:<\:0$$

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

For $$\displaystyle x > -1,\;\log_4(x+3) \:>\:0$$

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

Take anti-logs: . $$\displaystyle x+3 \,>\;\dfrac{x+1}{x+2}$$

. . which simplifies to: .$$\displaystyle x^2 + 4x + 5 \:>\:0\;\;[3]$$

This is an up-opening parabola whose vertex is at (-2,1).
. . Hence, the function is always positive.

The inequality is true for any $$\displaystyle x$$ in the domain: .$$\displaystyle (-1,\:\infty)$$

grgrsanjay