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Solve in x equation log(base 5)(5^(x+1)+5)*log(base 5)(5^x +1)=2

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minlux | Honors

Posted August 3, 2013 at 3:32 PM via web

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Solve in x equation log(base 5)(5^(x+1)+5)*log(base 5)(5^x +1)=2

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aruv | High School Teacher | Valedictorian

Posted August 3, 2013 at 3:54 PM (Answer #1)

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`log_5(5^(x+1)+5)xxlog_5(5^x+1)=2`

consider

`log_5(5^(x+1)+5)=log_5(5^x xx5+5)=log_5(5xx(5^x+1))`

`=log_5 5+log_5(5^x+1)=1+log_5(5^x+1)`

Let

`log_5(5^x+1)=y`

Thus

`log_5(5^(x+1)+5)xxlog_5(5^x+1)=`

`(1+log_5(5^x+1))xxlog_5(5^x+1)=(1+y)y=2`

`y+y^2=2`

`y^2+y-2=0`

`y^2+2y-y-2=0`

`(y+2)(y-1)=0`

`Thus`

`y=-2 and y=1`

i.e.

`log_5(5^x+1)=-2`

`5^(-2)=5^x+1`

`1/25=5^x+1`

`1=25xx5^x+25`

`1-25=5^(x+2)`

`-24=5^(x+2)`  which is not possible fr any value of x.

and

`log_5(5^x+1)=1`

`5^1=5^x+1`

`5-1=5^x`

`4=5^x`

`log(4)=log(5^x)`

`log(4)=xlog(5)`

`x=log(4)/log(5)`

`=.8614`

Thus x=.8614  ans.

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