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How to solve for x in log(8x)-log(1+x^1/2)=2

(common log)*

 Apr 2, 2015

Best Answer 

 #2
avatar+33654 
+10

In this case Melody, it might be better to let y = √x  (a positive number) when you get to 0.08x - 1 = √x  so that you have  0.08y2 - 1 = y.  Then when you solve for y, if you get a negative number you can discard it immediately:

 

0.08×y21=y{y=(5×3325)4y=(5×33+25)4}{y=0.9307033081725358y=13.4307033081725358}

 

Taking only the positive value we then have (approximately)

x=13.43072x=180.38370249

.

 Apr 2, 2015
 #1
avatar+118696 
+10

log(8x)log(1+x1/2)=2log8x1+x0.5=210log8x1+x0.5=1028x1+x0.5=1008x=100(1+x0.5)0.08x=1+x0.08x1=x0.0064x20.16x+1=x0.0064x21.16x+1=0

 

x is approx  0.866  or  180.384  but these must be checked. :)

 

0.0064×x21.16×x+1=0{x=(125×33725)8x=(125×33+725)8}{x=0.8662086478433022x=180.3837913521566978}

 

log(8x)-log(1+x^1/2)=2

check

log10(8×0.866208643)log10(1+0.8662086430.5)=0.5549969596406122

 

log10(8×180.383791352)log10(1+180.3837913520.5)=1.9999999999997984

 

Sox180.184

.
 Apr 2, 2015
 #2
avatar+33654 
+10
Best Answer

In this case Melody, it might be better to let y = √x  (a positive number) when you get to 0.08x - 1 = √x  so that you have  0.08y2 - 1 = y.  Then when you solve for y, if you get a negative number you can discard it immediately:

 

0.08×y21=y{y=(5×3325)4y=(5×33+25)4}{y=0.9307033081725358y=13.4307033081725358}

 

Taking only the positive value we then have (approximately)

x=13.43072x=180.38370249

.

Alan Apr 2, 2015
 #3
avatar+118696 
+5

Yes that is a much better idea - thanks Alan.      

 Apr 2, 2015

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