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avatar+597 

My second one for today:

 

Solve x:

 

2 Log x + 2 = log 900

 

I said:

 

\(Log X^2 + 2 = Log 30^2\)

\(2Log x - 2Log 30 = -2\)

\(2Log {x \over 30} = -2\)

\(60Logx = -60\)

\(Log x = -1\)

\(x=10^-1\)

\(x= {1 \over 10}\)

 

Is this correct?

 

wink

 Nov 13, 2019

Best Answer 

 #1
avatar+28250 
+3

You go wrong on your fourth line.    Check:  2log(1/10) +2 =0     log(900) =  2.954

 

\(2log_{10}x+2=log_{10}900\\ 2log_{10}x+2=log_{10}30^2\\2log_{10}x+2=2log_{10}30\\ log_{10}x+1=log_{10}30\\log_{10}x+log_{10}10=log_{10}30\\log_{10}(10x)=log_{10}30\\ 10x=30\\x = 3\)

.
 Nov 13, 2019
 #1
avatar+28250 
+3
Best Answer

You go wrong on your fourth line.    Check:  2log(1/10) +2 =0     log(900) =  2.954

 

\(2log_{10}x+2=log_{10}900\\ 2log_{10}x+2=log_{10}30^2\\2log_{10}x+2=2log_{10}30\\ log_{10}x+1=log_{10}30\\log_{10}x+log_{10}10=log_{10}30\\log_{10}(10x)=log_{10}30\\ 10x=30\\x = 3\)

Alan Nov 13, 2019
 #2
avatar+597 
+1

Alan,

 

thanks again...I understand my error...thanx a million!

juriemagic  Nov 13, 2019

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