logarithms

May 2010
143
15
chennai,tamil nadu
solve the eqn:

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

Soroban

MHF Hall of Honor
May 2006
12,028
6,341
Lexington, MA (USA)
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: .\(\displaystyle (-1,\:\infty)\)

 
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