How do you solve $10^{x + 2} - 12 = 22$ $10^(x+2) - 12 = 22$ and find any extraneous solutions?

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Answer 1 · 2016-07-22T10:18:07

The Soln. is $x = - 0.4685$ $x=-0.4685$

$10^{x + 2} - 12 = 22 \Rightarrow 10^{x + 2} = 12 + 22 = 34$ $10^(x+2)-12=22 rArr 10^(x+2)=12+22=34$

$\Rightarrow {log}_{10} 10^{x + 2} = {log}_{10} 34$ $rArrlog_(10)10^(x+2)=log_(10)34$

$\Rightarrow (x + 2) {log}_{10} 10 = 1.5315$ $rArr (x+2)log_(10)10=1.5315$ ..................[Using LogTable]

$\Rightarrow (x + 2) \cdot 1 = 1.5315$ $rArr (x+2)*1=1.5315$

$\Rightarrow x + 2 = 1.5315$ $rArr x+2=1.5315$

$\Rightarrow x = 1.5315 - 2$ $rArr x=1.5315-2$

$\Rightarrow x = - 0.4685$ $rArr x=-0.4685$ is the Soln.

How do you solve 10^(x+2) - 12 = 22 10x+2−12=2210^(x+2) - 12 = 22 and find any extraneous solutions?