Check for Extraneous Solutions

**paigesisco** · March 12th 2013, 09:08 AM

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.

**Plato** · March 12th 2013, 09:27 AM

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)

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

**MichaelLitzky** · March 12th 2013, 09:31 AM

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.

**Soroban** · March 12th 2013, 01:18 PM

Hello, paigesisco!

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

Note that the domain of the solution is: $\color{blue}{x > 6}.$
. . Also that: $\log_2(8) = 3$

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

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

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

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

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

But this is not in the domain.
It is an extraneous root.

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