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How do solve equation `25^(log_5 (lg x))=lgx-lg^2x+1?`  

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singup | Student, Grade 11 | (Level 2) eNoter

Posted September 5, 2012 at 12:37 PM via web

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How do solve equation `25^(log_5 (lg x))=lgx-lg^2x+1?`

 

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted September 5, 2012 at 1:28 PM (Answer #1)

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You should use the following logarithmic identity `n*log x = log x^n`  to solve `25^(log_5 (lg x))`  such that:

`25^(log_5 (lg x)) = 5^(2log_5 (lg x))`

You need to use the logarithmic identity `a^(log_a b) = b`  such that:

`5^(2log_5 (lg x)) = 5^(log_5 (lg x)^2) = lg^2 x`

Substituting `lg^2 x`  for `5^(2log_5 (lg x))`  yields:

`lg^2 x = lg x - lg^2 x + 1 => 2lg^2 x - lg x - 1 = 0`

You should come up with the following substitution `lg x = y`  such that:

`2y^2 - y - 1 = 0`

Using quadratic formula yields:

`y_(1,2) = (1+-sqrt(1+8))/4 => y_(1,2) = (1+-sqrt9)/4`

`y_(1,2) = (1+-3)/4 => y_1 = 1 ;y_2 = -1/2`

You need to solve for x the following equations `lg x = y_(1,2)`  such that:

`lg x = 1 => x = 10^1 => x = 10 `

`lg x = -1/2 => x = 10^(-1/2) => x = 1/(10^(1/2)) => x = 1/sqrt10`

Hence, evaluating the solutions to the given equation yields `x = 1/sqrt10`  and `x = 10` .

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