Check for Extraneous Solutions

My class is having a hard time understanding how to do these because we don't understand the Inverse Property solution in the problem

and example problem we are having trouble with is:

Log base[2](x-6)=3+log base[2](x-1)

A solution to this problem and extra advice would be nice.

Re: Check for Extraneous Solutions

Quote:

Originally Posted by

**paigesisco** My class is having a hard time understanding how to do these because we don't understand the Inverse Property solution in the problem

and example problem we are having trouble with is:

Log base[2](x-6)=3+log base[2](x-1)

$\displaystyle \\log_2(x-6)=3+\log_2(x-1)\\log_2\left(\frac{x-6}{x-1}\right)=3\\\left(\frac{x-6}{x-1}\right)=8$

Re: Check for Extraneous Solutions

And then to check for extraneous solutions: plug your x value back into the original equation (which of course we all do every time anyway <grin>). If you'd wind up taking the log of a negative number, then that solution doesn't work and is an extraneous solution.

Re: Check for Extraneous Solutions

Hello, paigesisco!

Quote:

$\displaystyle \text{Solve for }x\!:\;\;\log_2(x -6) \:=\:3 + \log_2(x-1)$

Note that the domain of the solution is: $\displaystyle \color{blue}{x > 6}.$

. . Also that: $\displaystyle \log_2(8) = 3$

We have: .$\displaystyle \log_2(x-6) \:=\:\log_2(8) + \log_2(x-1)$

. . . . . . . . $\displaystyle \log_2(x-6) \;=\;\log_2\big[8(x-1)\big]$

. . . . . . . . . . . . $\displaystyle x-6 \;=\;8x-8$

. . . . . . . . . . . . . $\displaystyle \text{-}7x \;=\;\text{-}2$

. . . . . . . . . . . . . . $\displaystyle x \;=\;\tfrac{2}{7}$

But this is not in the domain.

It is an extraneous root.